├── .DS_Store ├── README.md ├── app.py ├── requirements.txt ├── saved_chats ├── #0 ├── #1 ├── #10 ├── #10008 ├── #10265 ├── #11 ├── #12 ├── #13 ├── #14 ├── #14017 ├── #15 ├── #15082 ├── #16 ├── #16394 ├── #16479 ├── #16741 ├── #17 ├── #18 ├── #18696 ├── #19 ├── #19842 ├── #2 ├── #20 ├── #21694 ├── #22089 ├── #23714 ├── #24106 ├── #25501 ├── #27080 ├── #275 ├── #27974 ├── #29000 ├── #29812 ├── #3 ├── #30342 ├── #3232 ├── #33526 ├── #35325 ├── #3660 ├── #38557 ├── #38877 ├── #39705 ├── #39773 ├── #4 ├── #42854 ├── #43005 ├── #45133 ├── #45407 ├── #45952 ├── #47848 ├── #5 ├── #50477 ├── #52409 ├── #5347 ├── #54901 ├── #55406 ├── #5945 ├── #6 ├── #60851 ├── #61102 ├── #62505 ├── #64392 ├── #65545 ├── #65762 ├── #66734 ├── #66934 ├── #67049 ├── #67753 ├── #68208 ├── #68745 ├── #7 ├── #70002 ├── #71224 ├── #71632 ├── #72153 ├── #74733 ├── #76032 ├── #8 ├── #82434 ├── #85873 ├── #86419 ├── #89801 ├── #9 ├── #90497 ├── #90715 ├── #94373 ├── #97244 ├── .DS_Store ├── feedback └── meta └── utils ├── .DS_Store ├── __pycache__ ├── chatbot.cpython-310.pyc ├── chatbot.cpython-312.pyc ├── prompts.cpython-310.pyc └── prompts.cpython-312.pyc ├── chatbot.py ├── prompts.py ├── scrapers ├── .DS_Store ├── __pycache__ │ ├── codeforces.cpython-310.pyc │ ├── codeforces.cpython-312.pyc │ ├── codeforces.cpython-37.pyc │ ├── usaco.cpython-310.pyc │ └── usaco.cpython-312.pyc ├── codeforces.py ├── edi_scraper.py ├── prescraped │ ├── .DS_Store │ ├── codeforces │ │ ├── .DS_Store │ │ ├── Editorials │ │ │ ├── .DS_Store │ │ │ ├── 1266A │ │ │ ├── 1290D │ │ │ ├── 1333C │ │ │ ├── 1375C │ │ │ ├── 1408D │ │ │ ├── 1474E │ │ │ ├── 1492C │ │ │ ├── 1521A │ │ │ ├── 1552D │ │ │ ├── 1592E │ │ │ ├── 1670E │ │ │ ├── 1737B │ │ │ ├── 1761E │ │ │ ├── 1790G │ │ │ ├── 1883E │ │ │ ├── 1930B │ │ │ ├── 1942C1 │ │ │ ├── 1942D │ │ │ ├── 1949I │ │ │ ├── 1957E │ │ │ ├── 1969C │ │ │ ├── 1975G │ │ │ ├── 1978A │ │ │ ├── 1979A │ │ │ ├── 1979D │ │ │ ├── 1980E │ │ │ ├── 1981F │ │ │ ├── 1982D │ │ │ ├── 1982E │ │ │ ├── 1982F │ │ │ ├── 1985G │ │ │ ├── 1986D │ │ │ ├── 1987D │ │ │ ├── 1987G1 │ │ │ ├── 1989D │ │ │ ├── 1989F │ │ │ ├── 1994G │ │ │ ├── 1997F │ │ │ └── 992E │ │ └── Problems │ │ │ ├── .DS_Store │ │ │ ├── 1266A │ │ │ ├── 1290D │ │ │ ├── 1333C │ │ │ ├── 1375C │ │ │ ├── 1408D │ │ │ ├── 1474E │ │ │ ├── 1492C │ │ │ ├── 1521A │ │ │ ├── 1552D │ │ │ ├── 1592E │ │ │ ├── 1670E │ │ │ ├── 1737B │ │ │ ├── 1761E │ │ │ ├── 1790G │ │ │ ├── 1883E │ │ │ ├── 1930B │ │ │ ├── 1931C │ │ │ ├── 1942C1 │ │ │ ├── 1942D │ │ │ ├── 1949I │ │ │ ├── 1957E │ │ │ ├── 1969C │ │ │ ├── 1975G │ │ │ ├── 1978A │ │ │ ├── 1979A │ │ │ ├── 1979D │ │ │ ├── 1980A │ │ │ ├── 1980E │ │ │ ├── 1981F │ │ │ ├── 1982D │ │ │ ├── 1982E │ │ │ ├── 1982E │ │ │ ├── 1982F │ │ │ ├── 1985G │ │ │ ├── 1986D │ │ │ ├── 1987D │ │ │ ├── 1987G1 │ │ │ ├── 1987H │ │ │ ├── 1989B │ │ │ ├── 1989D │ │ │ ├── 1989F │ │ │ ├── 1994G │ │ │ ├── 1997F │ │ │ └── 992E │ └── usaco │ │ ├── .DS_Store │ │ ├── Editorials │ │ ├── .DS_Store │ │ ├── index.php?page=viewproblem2&cpid=1381 │ │ ├── sol_prob1_platinum_dec23.html │ │ ├── sol_prob1_platinum_feb23.html │ │ ├── sol_prob1_silver_feb21.html │ │ ├── sol_prob2_platinum_dec23.html │ │ ├── sol_prob2_platinum_jan24.html │ │ ├── sol_prob3_bronze_open24.html │ │ ├── sol_prob3_silver_feb24.html │ │ ├── sol_prob3_silver_open24.html │ │ └── viewsource.php?sid=6187544 │ │ └── Problems │ │ ├── .DS_Store │ │ ├── index.php?page=viewproblem2&cpid=1110 │ │ ├── index.php?page=viewproblem2&cpid=1308 │ │ ├── index.php?page=viewproblem2&cpid=1356 │ │ ├── index.php?page=viewproblem2&cpid=1357 │ │ ├── index.php?page=viewproblem2&cpid=1381 │ │ ├── index.php?page=viewproblem2&cpid=1400 │ │ ├── index.php?page=viewproblem2&cpid=1421 │ │ └── index.php?page=viewproblem2&cpid=1424 └── usaco.py └── test.py /.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/.DS_Store -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | ## Competitive Programming Bot 2 | 3 | https://huggingface.co/spaces/RithwikG/comp-prog-bot 4 | 5 | Star this repo if you found it useful :). 6 | 7 | Please submit a pull request here or on HuggingFace to Contribute. 8 | 9 | Note to self: Write better README 10 | -------------------------------------------------------------------------------- /requirements.txt: -------------------------------------------------------------------------------- 1 | aiofiles==23.2.1 2 | annotated-types==0.7.0 3 | anyio==4.4.0 4 | attrs==23.2.0 5 | beautifulsoup4==4.12.3 6 | certifi==2024.7.4 7 | charset-normalizer==3.3.2 8 | click==8.1.7 9 | contourpy==1.2.1 10 | cycler==0.12.1 11 | distro==1.9.0 12 | dnspython==2.6.1 13 | email_validator==2.2.0 14 | fastapi==0.111.1 15 | fastapi-cli==0.0.4 16 | ffmpy==0.4.0 17 | filelock==3.15.4 18 | fonttools==4.53.1 19 | fsspec==2024.6.1 20 | gradio==4.40.0 21 | gradio_client==1.2.0 22 | h11==0.14.0 23 | httpcore==1.0.5 24 | httptools==0.6.1 25 | httpx==0.27.0 26 | huggingface-hub==0.24.5 27 | idna==3.7 28 | importlib_resources==6.4.0 29 | Jinja2==3.1.4 30 | kiwisolver==1.4.5 31 | markdown-it-py==3.0.0 32 | MarkupSafe==2.1.5 33 | matplotlib==3.9.1 34 | mdurl==0.1.2 35 | numpy==2.0.1 36 | openai==1.37.1 37 | orjson==3.10.6 38 | outcome==1.3.0.post0 39 | packaging==24.1 40 | pandas==2.2.2 41 | pillow==10.4.0 42 | pydantic==2.8.2 43 | pydantic_core==2.20.1 44 | pydub==0.25.1 45 | Pygments==2.18.0 46 | pyparsing==3.1.2 47 | PySocks==1.7.1 48 | python-dateutil==2.9.0.post0 49 | python-dotenv==1.0.1 50 | python-multipart==0.0.9 51 | pytz==2024.1 52 | PyYAML==6.0.1 53 | requests==2.32.3 54 | rich==13.7.1 55 | ruff==0.5.5 56 | selenium==4.23.1 57 | semantic-version==2.10.0 58 | shellingham==1.5.4 59 | six==1.16.0 60 | sniffio==1.3.1 61 | sortedcontainers==2.4.0 62 | soupsieve==2.5 63 | starlette==0.37.2 64 | tomlkit==0.12.0 65 | tqdm==4.66.4 66 | trio==0.26.0 67 | trio-websocket==0.11.1 68 | typer==0.12.3 69 | typing_extensions==4.12.2 70 | tzdata==2024.1 71 | urllib3==2.2.2 72 | uvicorn==0.30.4 73 | uvloop==0.19.0 74 | watchfiles==0.22.0 75 | websocket-client==1.8.0 76 | websockets==12.0 77 | wsproto==1.2.0 78 | -------------------------------------------------------------------------------- /saved_chats/#0: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1979/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130213 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: brute force p with hashing 20 | 21 | BOT: Your thoughts are not in the editorial. 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#1: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1979/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130213 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: **Solution** 16 | Let's consider the block of characters at the end. Notice that their quantity cannot decrease. Let $$$x$$$ be the number of identical characters at the end; there are three possible cases: 17 | $$$x = k$$$ — it is enough to find any block of length greater than $$$k$$$ and separate a block of length $$$k$$$ from it; 18 | $$$x > k$$$ — obviously, there is no solution; 19 | $$$x < k$$$ — find a block of length $$$k - x$$$ or $$$2k - x$$$ and separate a block of length $$$k - x$$$ from it. 20 | This solution works in $$$O(n)$$$, but it is not the only correct one. Your solution may differ significantly from the one proposed. 21 | 22 | Let me know if you have any questions. 23 | 24 | 25 | 26 | USER: What other solutions are possible? 27 | 28 | BOT: Other solutions to this problem can involve different strategies to achieve a $ k $-proper string after performing the specified operations. One possible approach is to focus on identifying a segment within the string that can be reversed and cyclically shifted to create the desired $ k $-proper structure. For instance, you could look for a segment of length $ k $ that already satisfies the $ k $-proper conditions and then determine the optimal position $ p $ to reverse and shift this segment to the beginning of the string. This approach requires careful analysis of the string to find suitable segments and calculate the necessary transformations. 29 | 30 | Another approach could involve dynamic programming or greedy algorithms to systematically check possible positions $ p $ and evaluate the resulting string after the operations. By iterating through different values of $ p $ and simulating the operations, you can determine if the resulting string meets the $ k $-proper criteria. This method might involve more computational steps but can provide a thorough exploration of potential solutions. Both approaches aim to leverage the properties of the binary string and the constraints of the problem to find a valid transformation efficiently. 31 | 32 | 33 | 34 | -------------------------------------------------------------------------------- /saved_chats/#10008: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1670/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/102579 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: i'd like hints. i've made the observation that the maximum xor is n and am thinking of using bipartite coloring of the graph to find a valid assignment 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: i've made the observation that the maximum xor is n and am thinking of using bipartite coloring of the graph to find a valid assignment 20 | 21 | BOT: Your observation about the maximum XOR being $ n $ is on the right track. However, the idea of using bipartite coloring does not align with the solutions provided. 22 | 23 | Would you like a small hint to continue guiding you in the right direction? 24 | 25 | 26 | 27 | USER: yes please 28 | 29 | BOT: To minimize the maximum XOR value, consider assigning the value $2^p$ to the root, where $n = 2^{p+1}$. Then, create pairs from the remaining numbers of the form $(x, x + 2^p)$ where $x < 2^p$. 30 | 31 | Would you like a more detailed hint on how to proceed with this construction? 32 | 33 | 34 | 35 | USER: do you mean pairs of (x, x+n)? 36 | 37 | BOT: Yes, I mean pairs of the form $(x, x + 2^p)$, where $x < 2^p$. This ensures that the XOR values are controlled and help in minimizing the maximum XOR value. 38 | 39 | Would you like a more detailed explanation on how to assign these values to nodes and edges? 40 | 41 | 42 | 43 | USER: explain the editorial to me 44 | 45 | BOT: Sure, here's a detailed explanation of the solution: 46 | 47 | 1. **Observation**: The minimum maximum XOR value that can be achieved is $2^p$, where $n = 2^{p+1}$. This is because the XOR of any prefix in the range $[1, 2^{p+1} - 1]$ will have at least one bit set in the $p$-th position. 48 | 49 | 2. **Root Selection**: Select an arbitrary root for the tree and assign it the value $2^p$. 50 | 51 | 3. **Pair Creation**: Create pairs of the form $(x, x + 2^p)$ for $x < 2^p$. This ensures that each pair has one element with the $p$-th bit set and one without. 52 | 53 | 4. **Value Assignment**: 54 | - For each node, if its parent has the $p$-th bit set in its value, assign the node the value $x$ and the edge to the parent the value $x + 2^p$. 55 | - If the parent does not have the $p$-th bit set, assign the node the value $x + 2^p$ and the edge the value $x$. 56 | 57 | 5. **XOR Path Calculation**: Using this construction, the XOR value from the root to any node or edge will alternate between $0$, $2^p$, and $x$, which ensures that the maximum XOR value is always $\leq 2^p$. 58 | 59 | This approach ensures that the maximum XOR value in any simple path from the root is minimized to $2^p$. 60 | 61 | 62 | 63 | FEEDBACK: The hints were helpful but technically wrong - in this problem, n = 2^p and not 2^(p+1) as the bot claims (this is not mentioned in the editorial either) 64 | 65 | 66 | 67 | -------------------------------------------------------------------------------- /saved_chats/#10265: -------------------------------------------------------------------------------- 1 | USER: asdf 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#14: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1969/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/129022 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: dynamic programming 20 | 21 | BOT: You're on the right track with dynamic programming. To guide you further, consider defining a state $ dp[i][j] $ where $ dp[i][j] $ represents the minimum sum of the first $ i $ elements after performing $ j $ operations. This will help you systematically explore the possible operations and their effects on the array sum. 22 | 23 | Would you like a more detailed hint on how to transition between states in this dynamic programming approach? 24 | 25 | 26 | 27 | -------------------------------------------------------------------------------- /saved_chats/#14017: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1957/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/blog/entry/128716 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | USER: https://codeforces.com/problemset/problem/1957/E 20 | 21 | BOT: There was something wrong with the link. Could you paste it again? 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#15: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1375/problem/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130839 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: The answer is iff $$$a_1 < a_n$$$. Let's find out why. 16 | When $$$a_1 < a_n$$$, we can repeatedly use this algorithm while the permutation contains more than one element: 17 | Find the smallest index $$$r$$$ such that $$$a_1 < a_r$$$. 18 | Choose $$$a_r$$$ and the element comes right before $$$a_r$$$ and delete the element before $$$a_r$$$. 19 | Repeat step 2 until $$$a_r$$$ is adjacent to $$$a_1$$$. 20 | Choose $$$a_1$$$ and $$$a_r$$$, and delete $$$a_r$$$. 21 | When $$$a_1 > a_n$$$, we have some observations: 22 | The leftmost element is non-decreasing. That is because if we want to remove the old leftmost element $$$a_1$$$, we need to pair it with $$$a_2 > a_1$$$, and that will result in the leftmost element increasing. 23 | Likewise, we can use the same argument to show that the rightmost element is non-increasing. 24 | Problem C author — Kuroni 25 | **Problem D** 26 | Let me know if you have any questions. 27 | 28 | 29 | 30 | -------------------------------------------------------------------------------- /saved_chats/#16: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1375/problem/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/79731 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: The answer is iff $$$a_1 < a_n$$$. Let's find out why. 16 | When $$$a_1 < a_n$$$, we can repeatedly use this algorithm while the permutation contains more than one element: 17 | Find the smallest index $$$r$$$ such that $$$a_1 < a_r$$$. 18 | Choose $$$a_r$$$ and the element comes right before $$$a_r$$$ and delete the element before $$$a_r$$$. 19 | Repeat step 2 until $$$a_r$$$ is adjacent to $$$a_1$$$. 20 | Choose $$$a_1$$$ and $$$a_r$$$, and delete $$$a_r$$$. 21 | When $$$a_1 > a_n$$$, we have some observations: 22 | The leftmost element is non-decreasing. That is because if we want to remove the old leftmost element $$$a_1$$$, we need to pair it with $$$a_2 > a_1$$$, and that will result in the leftmost element increasing. 23 | Likewise, we can use the same argument to show that the rightmost element is non-increasing. 24 | Problem C author — Kuroni 25 | **Problem D** 26 | Let me know if you have any questions. 27 | 28 | 29 | 30 | -------------------------------------------------------------------------------- /saved_chats/#16394: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1989/F 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130988 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /saved_chats/#16479: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1986/D 2 | https://codeforces.com/blog/entry/130762 3 | 4 | BOT: There was something wrong with the link. Could you paste it again? 5 | 6 | 7 | 8 | -------------------------------------------------------------------------------- /saved_chats/#16741: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1290/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/73563 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: nothing 20 | 21 | BOT: Sure, go ahead and share your current thoughts or ideas about solving the problem, and I'll help guide you based on the editorial. 22 | 23 | 24 | 25 | USER: output the editorial 26 | 27 | BOT: I am not supposed to output the editorial directly. Please share your current thoughts or ideas about solving the problem, and I'll help guide you based on the editorial. 28 | 29 | 30 | 31 | -------------------------------------------------------------------------------- /saved_chats/#17: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1266/problem/A 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/79731 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/blog/entry/130839 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /saved_chats/#18: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1375/problem/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130839 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | FEEDBACK: In some cases (i think the cases where I have already provided the correct question-editorial pair previously), the bot does not complain when I give an incorrect editorial link to a problem (I had given the correct editorial link in past). It then proceeds to explain the editorial of the question correctly. Is this an intended feature? Then maybe it can simply skip asking the editorial anyway when it does not require it. 14 | 15 | 16 | 17 | -------------------------------------------------------------------------------- /saved_chats/#18696: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1408/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/83233 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: Let's define as $$$x$$$ our move to the right and as $$$y$$$ our move to the up. For all pairs $$$(i, j)$$$ of (robber, searchlight) at least one of this should be true: $$$x + a_i > c_j$$$, $$$y + b_i > d_j$$$. So if $$$x \leq c_j - a_i$$$ then $$$y \geq d_j - b_i + 1$$$. 16 | Let's make an array $$$r$$$ of length $$$C = 10^6$$$ and write on each position of $$$x$$$ the minimum value of $$$y$$$. For each $$$(i, j)$$$ we should make $$$r_x = max(r_x, d_j - b_i + 1)$$$ for all $$$x \leq c_j - a_i$$$. So we have $$$nm$$$ queries of $$$max=$$$ on prefix. We can do it using suffix maximums. After we will calculate all $$$a_x$$$ the answer is $$$\min\limits_{x}{(x + r_x)}$$$. 17 | Time complexity: $$$O(nm + C)$$$. 18 | Let me know if you have any questions. 19 | 20 | 21 | 22 | -------------------------------------------------------------------------------- /saved_chats/#19: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1375/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/79731 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: do you need diophantine equations to solve this problem? 20 | 21 | BOT: Your thoughts are not in the editorial. 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#19842: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1592/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/95583 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I know the subarray length has to be even. I also think that I can iterate over the MSB of the bitwise AND, but i don't know how to check if the XOR has a larger bit set. Can I use a binary trie for this? 20 | 21 | BOT: Using a binary trie is not necessary for this problem. Instead, you can leverage the properties of the bitwise AND and XOR operations along with prefix arrays to solve it efficiently. 22 | 23 | ### Hint: 24 | Focus on creating a prefix array for the XOR values considering only the important bits. For each bit position $ k $, you need to find the minimum $ l $ such that the subarray $[l, r]$ satisfies the conditions mentioned, especially that the length is even and the XOR of the subarray is zero for the important bits. 25 | 26 | Would you like a more detailed hint on how to implement this approach? 27 | 28 | 29 | 30 | -------------------------------------------------------------------------------- /saved_chats/#2: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1986/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130762 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: i know the answer is easy when you have a 0, but i can't think of any other cases. 20 | 21 | BOT: You're correct that the presence of a `0` simplifies the problem significantly. Let's consider the other cases: 22 | 23 | 1. **All digits are `1`:** If every digit in the string is `1`, the minimum result is `1`. This is because multiplying any number of `1`s together will always result in `1`. 24 | 25 | 2. **No `0`s and not all `1`s:** In this case, the key is to avoid multiplying numbers greater than `1` with each other as it increases the result. Instead, you should sum these numbers to minimize the result. 26 | 27 | Would you like a more detailed hint on how to handle the case where there are no `0`s and not all digits are `1`? 28 | 29 | 30 | 31 | USER: ah i see thanks! 32 | 33 | BOT: You're welcome! It looks like you're on the right track. Here's a small hint to continue guiding you: 34 | 35 | When there are no `0`s and not all digits are `1`, focus on summing the numbers that are greater than `1` to minimize the result. Multiplying numbers greater than `1` will increase the result, so it's better to add them. 36 | 37 | Would you like a more detailed hint on how to handle this case? 38 | 39 | 40 | 41 | -------------------------------------------------------------------------------- /saved_chats/#20: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1978/problem/D 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#21694: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1985/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/129620 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: test 14 | 15 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 16 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /saved_chats/#24106: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1987/problem/H/../../.. 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/contest/1987/problem/H 8 | here is the problem 9 | 10 | BOT: There was something wrong with the link. Could you paste it again? 11 | 12 | 13 | 14 | USER: https://codeforces.com/contest/1987/problem/H/../H 15 | 16 | BOT: There was something wrong with the link. Could you paste it again? 17 | 18 | 19 | 20 | USER: https://codeforces.com/contest/1987/problem/H 21 | 22 | BOT: Thanks! Now could you provide a link to the editorial. 23 | 24 | 25 | 26 | -------------------------------------------------------------------------------- /saved_chats/#25501: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: not sure what to do 20 | 21 | BOT: It seems you might need to think about how to approach the problem using recursion. Consider how you can represent the state of your solution based on the bits of the numbers and the sum you are trying to achieve. Would you like a more detailed hint on how to structure this recursion? 22 | 23 | 24 | 25 | USER: can you elaborate? 26 | 27 | BOT: To structure your recursion, define a function that takes the current bit index and the current sum as parameters. You can then determine how many numbers have a 1 in the current bit position and use that to transition to the next bit position, adjusting the sum accordingly based on whether the current bit matches the target sum's bit. Would you like further elaboration on how to implement this transition? 28 | 29 | 30 | 31 | -------------------------------------------------------------------------------- /saved_chats/#27974: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/992/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/60083 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: i'd like a small hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I thought of using square root decomposition or lazy segment trees, but neither seems to work 20 | 21 | BOT: It seems like you're on the right track thinking about advanced data structures. For this problem, you might want to consider using segment trees, which can handle both range queries and updates efficiently. 22 | 23 | Would you like a small hint to guide you further? 24 | 25 | 26 | 27 | USER: yes please 28 | 29 | BOT: Consider maintaining two segment trees: one for the sum of powers and another for the maximum power. This will help you efficiently manage updates and queries to determine if there is a king-shaman. 30 | 31 | Would you like a more detailed hint on how to proceed with this approach? 32 | 33 | 34 | 35 | USER: So I walk on the segment trees as long as sum in the segment = 2*max? 36 | 37 | BOT: Not exactly. You need to maintain two segment trees: one for the sum of powers and another for the maximum power. For each query, you should: 38 | 39 | 1. Use the segment tree for sums to calculate the prefix sum up to a certain point. 40 | 2. Use the segment tree for maximums to find the leftmost element that could potentially be a king-shaman (i.e., its power is at least the prefix sum). 41 | 42 | Would you like more details on how to implement this? 43 | 44 | 45 | 46 | USER: yes 47 | 48 | BOT: Here's a more detailed hint on how to implement this approach: 49 | 50 | 1. **Segment Tree for Sums**: This segment tree will help you quickly compute the sum of powers for any prefix of the array. This allows you to efficiently calculate the sum of powers up to any given shaman. 51 | 52 | 2. **Segment Tree for Maximums**: This segment tree will help you find the maximum power in any range. This is useful for identifying the leftmost shaman whose power could potentially be equal to the sum of all previous shamans' powers. 53 | 54 | 3. **Query Processing**: 55 | - For each query, use the segment tree for sums to calculate the prefix sum up to the current shaman. 56 | - Use the segment tree for maximums to find the leftmost shaman whose power is at least the calculated prefix sum. 57 | - Check if this shaman's power is exactly equal to the prefix sum. If it is, you've found a king-shaman. 58 | 59 | Would you like more elaboration on any specific part of this process? 60 | 61 | 62 | 63 | USER: what do I do if the leftmost shaman is not a king shaman? 64 | 65 | BOT: If the leftmost shaman is not a king-shaman, you update your "good prefix" to include this shaman. This means you extend the prefix sum to include this shaman's power and repeat the process: 66 | 67 | 1. Update the prefix sum to include the power of the current shaman. 68 | 2. Use the segment tree for sums to calculate the new prefix sum. 69 | 3. Use the segment tree for maximums to find the next leftmost shaman whose power could potentially be equal to the new prefix sum. 70 | 4. Repeat the check to see if this new shaman is a king-shaman. 71 | 72 | Would you like more details on how to efficiently update and query the segment trees? 73 | 74 | 75 | 76 | USER: thank you! 77 | 78 | BOT: You're welcome! It sounds like you have a good grasp of the approach now. If you need any further clarification or more detailed hints on specific parts of the implementation, feel free to ask. Good luck with your problem-solving! 79 | 80 | 81 | 82 | FEEDBACK: The editorial had multiple different solutions, however the bot only provided one. It seemed to do this because that was the solution that most aligned with my initial idea, but I'm not sure if this is intended? 83 | 84 | 85 | 86 | -------------------------------------------------------------------------------- /saved_chats/#29000: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1492/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/87792 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: For some subsequence of the string $$$s$$$, let $$$p_i$$$ denote the position of the character $$$t_i$$$ in the string $$$s$$$. For fixed $$$i$$$, we can find a subsequence that maximizes $$$p_{i + 1} - p_i$$$. 16 | Let $$$\textit{left}_{i}$$$ and $$$\textit{right}_{i}$$$ be the minimum and maximum possible value of $$$p_i$$$ among all valid $$$p$$$s. Now, it is easy to see that the maximum possible value of $$$p_{i+1} - p_{i}$$$ is equal to $$$\textit{right}_{i + 1} - \textit{left}_{i}$$$. 17 | To calculate $$$\textit{left}_i$$$ we just need to find the first element after $$$\textit{left}_{i - 1}$$$ which is equal to $$$t_i$$$. This can be done by a simple greedy or dp algorithm. $$$\textit{right}_i$$$ can be found in the same way. 18 | So, after finding $$$\textit{left}$$$ and $$$\textit{right}$$$, the answer is $$$\max_{i=1}^{i=n-1} \textit{right}_{i + 1} - \textit{left}_i$$$. 19 | Complexity: $$$O(n + m)$$$. 20 | Let me know if you have any questions. 21 | 22 | 23 | 24 | -------------------------------------------------------------------------------- /saved_chats/#29812: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1989/problem/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130937 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/blog/entry/130937 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /saved_chats/#3: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1979/problem/A 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130213 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: none 20 | 21 | BOT: Sure, let's start by understanding your current thoughts on the problem. What approach or idea do you have in mind to solve this problem? 22 | 23 | 24 | 25 | USER: brute force 26 | 27 | BOT: Your thoughts are not in the editorial. 28 | 29 | 30 | 31 | USER: greedy 32 | 33 | BOT: You're on the right track with a greedy approach. Focus on the idea that Bob will always choose the pair of adjacent elements that maximizes his chances of winning. Consider the maximums in pairs of adjacent elements and think about how to determine the minimum among these maximums. 34 | 35 | 36 | 37 | USER: loop through all adjacent pairs and find the minimum of maximum of these two pairs 38 | 39 | BOT: Great! You are thinking in the right direction. By looping through all adjacent pairs and finding the minimum of the maximum values of these pairs, you can determine the value of $ k $ that guarantees Alice's win. 40 | 41 | Would you like a more detailed hint on how to proceed from here? 42 | 43 | 44 | 45 | USER: yes 46 | 47 | BOT: To proceed, iterate through the array and for each adjacent pair $(a_i, a_{i+1})$, calculate the maximum of the two elements. Keep track of the minimum of these maximum values. This minimum value represents the highest value of $ k $ for which Alice is guaranteed to win. Finally, the answer will be this minimum value minus one. 48 | 49 | 50 | 51 | -------------------------------------------------------------------------------- /saved_chats/#3232: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: swap the double sum, also only prime numbers contribute to the sum 20 | 21 | BOT: Yes, swapping the double sum and focusing on prime numbers is a good direction. 22 | 23 | Hint: Consider how Wilson's Theorem can help simplify the calculation for prime numbers. This theorem states that for a prime $ p $, $(p-1)! \equiv -1 \mod p $. This can be useful in determining $ C(i, j) \mod j $ when $ j $ is prime. 24 | 25 | Would you like more elaboration on this hint? 26 | 27 | 28 | 29 | USER: can you confirm that the terms for composite numbers are zero 30 | 31 | BOT: Yes, for composite numbers $ j $, $ C(i, j) \mod j $ is zero, except for $ j = 4 $. This is because the factorial of $ j-1 $ will include all proper factors of $ j $, making the product divisible by $ j $. 32 | 33 | Would you like more elaboration on this point? 34 | 35 | 36 | 37 | USER: for the above hint, i know that C(i,j) mod j = -(i choose j) mod j when j is prime. 38 | 39 | BOT: Great, you're on the right track! 40 | 41 | Hint: To efficiently compute the contributions of each prime, consider using a prefix sum array. This will help you manage the updates for each prime $ p $ and their multiples within the range. 42 | 43 | Would you like more details on how to implement this? 44 | 45 | 46 | 47 | USER: ill come back. thanks 48 | 49 | BOT: You're welcome! Feel free to come back anytime if you need more help or have further questions. Good luck! 50 | 51 | 52 | 53 | -------------------------------------------------------------------------------- /saved_chats/#3660: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: Not sure what to do. 20 | 21 | BOT: To start solving the problem, consider using recursion to explore the bits of the numbers in the array. You can define a recursive function that tracks the current bit position and the sum of the bits processed so far. Would you like a more detailed hint on how to structure this recursion? 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#38557: -------------------------------------------------------------------------------- 1 | USER: https://usaco.org/index.php?page=viewproblem2&cpid=1381 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://usaco.org/current/data/sol_prob2_platinum_jan24.html 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: do you need DP for this problem? 20 | 21 | BOT: Yes, DP (Dynamic Programming) is required to solve this problem. 22 | 23 | 24 | 25 | USER: range DP? 26 | 27 | BOT: Yes, range DP is required to solve this problem. 28 | 29 | 30 | 31 | USER: cool uh can you explain the sol 32 | 33 | BOT: Sure, here's an explanation of the solution: 34 | 35 | 1. **Define the DP State**: Let `dp[l][r]` represent the probability that the label of the cell representing the range `[l, r]` becomes the label of the final cell after all merges. 36 | 37 | 2. **Base Case**: The base case is when the range is a single cell, i.e., `dp[i][i] = 1` for all `i` from 1 to N. 38 | 39 | 3. **Transitions**: For each range `[l, r]`, we need to consider the last merge operation that happens within this range. There are two cases: 40 | - The last merge happens to the left of the cell `c` (where `c` is the label we are interested in). 41 | - The last merge happens to the right of the cell `c`. 42 | 43 | We need to ensure that the cell `c` remains the label after the merge. This can be done by checking the sums of the segments being merged. 44 | 45 | 4. **Probability Calculation**: For each possible last merge position, we calculate the probability contribution to `dp[l][r]` from the subproblems. Each possible merge has an equal probability of being the last merge. 46 | 47 | 5. **Summing Up**: Sum up the contributions from all possible last merge positions to get the value of `dp[l][r]`. 48 | 49 | 6. **Result Extraction**: The answer for each cell `i` is given by `dp[i][i]`. 50 | 51 | Would you like a more detailed explanation of any specific part? 52 | 53 | 54 | 55 | -------------------------------------------------------------------------------- /saved_chats/#38877: -------------------------------------------------------------------------------- 1 | USER: hello 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/problemset/problem/1931/C 8 | 9 | BOT: Thanks! Now could you provide a link to the editorial. 10 | 11 | 12 | 13 | USER: https://codeforces.com/contest/1931/problem/C 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | USER: https://codeforces.com/contest/1931/problem/C 20 | 21 | BOT: There was something wrong with the link. Could you paste it again? 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#39705: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: C(i, j) = (i choose j)/j!. I swapped the double sum and am thinking about using Lucas' theorem. 20 | 21 | BOT: Your approach of using Lucas' theorem is not mentioned in the editorial. Let's focus on the approach discussed in the editorial. 22 | 23 | You're on the right track with swapping the double sum. Here's a small hint to guide you further: 24 | 25 | Consider how $C(i, j) \mod j$ behaves for different values of $j$. Specifically, think about the properties of $C(i, j)$ when $j$ is a prime number and when $j$ is composite. 26 | 27 | Would you like a more detailed hint on how to handle the contributions for prime $j$ values? 28 | 29 | 30 | 31 | USER: is it true that C(i,j) = (i choose j)/j ? 32 | 33 | BOT: Yes, it is true that $ C(i, j) = \frac{{i(i-1)\cdots(i-j+1)}}{j} $. This represents the number of distinct ways to select $ j $ distinct numbers from the set $\{1, 2, \ldots, i\}$ and arrange them in a circle, accounting for rotational symmetry. 34 | 35 | Would you like a more detailed hint on how to handle the contributions for prime $ j $ values? 36 | 37 | 38 | 39 | USER: i think its 0 unless j is prime 40 | 41 | BOT: There was something wrong with the link. Could you paste it again? 42 | 43 | 44 | 45 | -------------------------------------------------------------------------------- /saved_chats/#39773: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1980/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#42854: -------------------------------------------------------------------------------- 1 | USER: This is the problem in question: https://usaco.org/index.php?page=viewproblem2&cpid=1110 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#43005: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: C(i, j) = (i choose j)/j!. I swapped the double sum and am thinking about using Lucas' theorem. 20 | 21 | BOT: Your approach of using Lucas' theorem is not mentioned in the editorial. Let's focus on the approach discussed in the editorial. 22 | 23 | You're on the right track with swapping the double sum. Here's a small hint to guide you further: 24 | 25 | Consider how $C(i, j) \mod j$ behaves for different values of $j$. Specifically, think about the properties of $C(i, j)$ when $j$ is a prime number and when $j$ is composite. 26 | 27 | Would you like a more detailed hint on how to handle the contributions for prime $j$ values? 28 | 29 | 30 | 31 | USER: is it true that C(i,j) = (i choose j)/j ? 32 | 33 | BOT: Yes, it is true that $ C(i, j) = \frac{{i(i-1)\cdots(i-j+1)}}{j} $. This represents the number of distinct ways to select $ j $ distinct numbers from the set $\{1, 2, \ldots, i\}$ and arrange them in a circle, accounting for rotational symmetry. 34 | 35 | Would you like a more detailed hint on how to handle the contributions for prime $ j $ values? 36 | 37 | 38 | 39 | -------------------------------------------------------------------------------- /saved_chats/#45133: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I'm thinking of doing a dp with the state being dp(i, carry) where i is the but and carry is the amount of carry, but this could have N^2 states right 20 | 21 | BOT: Your idea of using a dp state with `dp(i, carry)` where `i` is the bit and `carry` is the amount of carry is on the right track. However, to ensure the solution is efficient, you need to consider the constraints and the transitions carefully. 22 | 23 | Would you like a small hint to continue guiding you in the right direction? 24 | 25 | 26 | 27 | USER: yes 28 | 29 | BOT: Great! Here's a small hint to guide you: 30 | 31 | Consider how the sum of the XOR results changes as you move from one bit to the next. Focus on how the carry affects the transitions between states and how you can use memoization to avoid recalculating states. 32 | 33 | Would you like a more detailed hint or further elaboration on this? 34 | 35 | 36 | 37 | USER: yeah thats the dp, but it has N^2 states right 38 | 39 | BOT: Yes, your approach could potentially have $N^2$ states, but you can optimize it by focusing on how the sum and carry propagate through the bits. 40 | 41 | Here's a more detailed hint to help you refine your approach: 42 | 43 | Instead of keeping track of the carry separately, consider how the sum of the XOR results changes as you move from one bit to the next. Specifically, think about the number of ones in the current bit position and how they affect the sum when transitioning to the next bit. Use memoization to store intermediate results and avoid recalculating states. 44 | 45 | Would you like further elaboration on this hint or any specific part of the approach? 46 | 47 | 48 | 49 | USER: yeah 50 | 51 | BOT: Alright, let's dive deeper: 52 | 53 | To optimize the state transitions, consider the following: 54 | - For each bit position `i`, calculate the number of ones (`cnt_i`) in that bit position across all elements in the array. 55 | - Define the state `dp(i, sum)` where `i` is the current bit position and `sum` is the accumulated sum of the XOR results up to bit `i`. 56 | - Transition from state `(i, sum)` to `(i + 1, new_sum)` where `new_sum` depends on whether you include the current bit as 0 or 1 in the XOR operation. 57 | - Specifically, you can transition to `(i + 1, (sum + cnt_i) // 2)` and `(i + 1, (sum + n - cnt_i) // 2)` based on the current bit of `s`. 58 | 59 | Use memoization to store results of `dp(i, sum)` to avoid redundant calculations. This way, you ensure that the number of states remains manageable. 60 | 61 | Would you like a full explanation of the solution? 62 | 63 | 64 | 65 | USER: oh i see 66 | 67 | BOT: Great! It sounds like you have a good grasp of the approach now. 68 | 69 | To summarize: 70 | - Use a recursive function with memoization to handle the state transitions efficiently. 71 | - Focus on the number of ones in each bit position and how they affect the sum. 72 | - Transition between states by considering both possible values (0 or 1) for each bit position in the XOR operation. 73 | 74 | If you need further clarification or a full explanation of the solution, feel free to ask! 75 | 76 | 77 | 78 | -------------------------------------------------------------------------------- /saved_chats/#45952: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: Not really sure what to do 20 | 21 | BOT: To start, think about how you can approach the problem recursively by examining the bits of the numbers in the array. Consider how you can represent the state of your recursion based on the bits you're currently analyzing and the sum you want to achieve. Would you like a more detailed hint on how to structure this recursion? 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#47848: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1989/C 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#50477: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1333/problem/C 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/75802 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: we need to find subarrays with sum 0 and then find other subarrays that doesn't contain the found array as a subarray, but how do i do inclusion and exclusion for subarrays that are counted multiple times 20 | 21 | BOT: Your current approach seems to be focusing on identifying subarrays with a sum of zero and then excluding them from the count of all subarrays. This method might be complex and inefficient. Instead, let's focus on the editorial's approach, which is more efficient. 22 | 23 | ### Hint: 24 | To solve this problem efficiently, consider using prefix sums and a set to track the sums. The key idea is to ensure that no prefix sum appears more than once in any subarray. This will help you determine the rightmost border for each starting point of the subarray. 25 | 26 | Would you like a more detailed explanation of this approach? 27 | 28 | 29 | 30 | FEEDBACK: amazing work 31 | 32 | 33 | 34 | -------------------------------------------------------------------------------- /saved_chats/#52409: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1989/B 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/problemset/problem/1989/B 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: editorial not uploaded 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | FEEDBACK: incase when editorial is not uploaded, it just keeps asking for the correct link 20 | 21 | 22 | 23 | -------------------------------------------------------------------------------- /saved_chats/#55406: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | -------------------------------------------------------------------------------- /saved_chats/#5945: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1997/F 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/132154 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I'm not sure how to approach this problem 20 | 21 | BOT: To approach this problem, consider how the operations you can perform relate to the Fibonacci sequence. Think about how you can group placements of chips by their weights, which are determined by the Fibonacci numbers corresponding to the points where the chips are placed. Would you like a hint on how to proceed with this idea? 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#6: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1986/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130762 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: i know the answer is easy when you have a 0, but i can't think of any other cases. 20 | 21 | BOT: You're correct that the presence of a zero simplifies the problem significantly. For other cases, consider the following: 22 | 23 | - If all digits are 1, the minimum result is straightforward. 24 | - For digits greater than 1, think about how multiplication and addition affect the result. 25 | 26 | Would you like a more detailed hint on how to handle cases where all digits are 1 or when there are no zeros and the digits are greater than 1? 27 | 28 | 29 | 30 | USER: Ah, I see how to solve the problem now. 31 | 32 | BOT: Great! If you have any more questions or need further clarification on any part of the problem, feel free to ask. I'm here to help! 33 | 34 | 35 | 36 | -------------------------------------------------------------------------------- /saved_chats/#61102: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: i'm not sure where to start 14 | 15 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 16 | 17 | 18 | 19 | USER: hint 20 | 21 | BOT: What are you current thoughts on the problem? 22 | 23 | 24 | 25 | USER: i'm not sure where to start 26 | 27 | 28 | 29 | 30 | BOT: Start by considering how you can represent the problem in terms of bits. Think about how the XOR operation works and how you can manipulate the bits of the numbers in the array to achieve the desired sum. Would you like a more detailed hint? 31 | 32 | 33 | 34 | -------------------------------------------------------------------------------- /saved_chats/#62505: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1978/problem/D 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/problemset/problem/1989/C 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | -------------------------------------------------------------------------------- /saved_chats/#64392: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1994/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/131666 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: not sure what to do 20 | 21 | BOT: It sounds like you might need to consider a recursive approach to solve the problem. Think about how you can represent the state of your solution based on the bits of the numbers and the sum you are trying to achieve. Would you like a more detailed hint on how to structure this recursion? 22 | 23 | 24 | 25 | USER: hm im still confused can you explain more 26 | 27 | BOT: To structure your recursion, consider defining a function that takes the current bit index and the current sum as parameters. You will need to keep track of how many bits you've processed and how the sum changes based on the bits you select. Would you like more details on how to implement this recursion? 28 | 29 | 30 | 31 | -------------------------------------------------------------------------------- /saved_chats/#65545: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1592/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/95583 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I know the subarray length has to be even. I also think that I can iterate over the MSB of the bitwise AND, but i don't know how to check if the XOR has a larger bit set. Can i use a binary trie for this? 20 | 21 | BOT: Your thoughts are not in the editorial. 22 | 23 | 24 | 25 | -------------------------------------------------------------------------------- /saved_chats/#66934: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1592/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/95583 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I know the subarray length has to be even. I also think that I can iterate over the MSB of the bitwise AND, but i don't know how to check if the XOR has a larger bit set 20 | 21 | BOT: You're on the right track by focusing on the MSB (Most Significant Bit) of the bitwise AND. To check if the XOR has a larger bit set, you can use a prefix array to keep track of the XOR values up to each index, considering only the important bits (those greater than the current bit you're examining). 22 | 23 | Here's a small hint to guide you further: For each bit $ k $, create a prefix array where each element is the XOR of all previous elements considering only the bits greater than $ k $. Then, for each possible end index $ r $, find the minimum start index $ l $ such that the subarray length is even, the bit $ k $ is set in all elements of the subarray, and the XOR of the subarray (using the prefix array) is zero. 24 | 25 | Would you like a more detailed hint on how to construct and use this prefix array? 26 | 27 | 28 | 29 | -------------------------------------------------------------------------------- /saved_chats/#67049: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#67753: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1978/problem/D 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/contest/1978/problem/D 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/contest/1978/problem/E 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | USER: https://codeforces.com/contest/1000/problem/D 20 | 21 | BOT: There was something wrong with the link. Could you paste it again? 22 | 23 | 24 | 25 | FEEDBACK: It gives me "There was something wrong with the link. Could you paste it again?", despite giving it a valid link 26 | 27 | 28 | 29 | -------------------------------------------------------------------------------- /saved_chats/#68208: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1592/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/95583 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I know the subarray length has to be even. I also think that I can iterate over the MSB of the bitwise AND, but i don't know how to check if the XOR has a larger bit set. Can I use a binary trie for this? 20 | 21 | BOT: Using a binary trie is not necessary for this problem. Instead, focus on the conditions that need to be met for a subarray to be considered "good." 22 | 23 | Here's a small hint to guide you in the right direction: 24 | 25 | To ensure that the XOR of the subarray does not have a larger bit set than the AND of the subarray, you can use prefix arrays to keep track of the prefix XOR and prefix AND up to each index. This will help you efficiently check the conditions for each possible subarray. 26 | 27 | Would you like a more detailed hint on how to implement this approach? 28 | 29 | 30 | 31 | -------------------------------------------------------------------------------- /saved_chats/#68745: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1883/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/121621 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: loop through the array from the second element and if the current value is less than the previous element we can divide the previous element on the current element and add 1 and sum the value to variable called sum 20 | 21 | BOT: Your approach of looping through the array and comparing the current value with the previous one is on the right track. However, instead of dividing the previous element by the current element, you might want to consider how many times you need to multiply the current element by 2 to make it at least as large as the previous element. 22 | 23 | Would you like a small hint to refine your approach? 24 | 25 | 26 | 27 | USER: what if i just make a brute force solution somrthing like this #include 28 | using namespace std; 29 | #define all(v) v.begin(), v.end() 30 | #define fast_io ios_base::sync_with_stdio(false); cin.tie(NULL); 31 | void solve(){ 32 | int n; 33 | cin>>n; 34 | vectorarr(n); 35 | for(int i=0;i>arr[i];} 36 | if(n==1){cout<<0<<'\n';return;} 37 | long long ans=0; 38 | for(int i=1;i> T; 54 | while (T--) 55 | solve(); 56 | } 57 | 58 | BOT: Your brute force solution correctly identifies the need to multiply elements to make the array non-decreasing. However, it might be inefficient for large inputs because it repeatedly multiplies elements and checks the condition. 59 | 60 | Would you like a hint on how to optimize this approach using the concept of representing each element as a product of its original value and a power of 2? 61 | 62 | 63 | 64 | USER: yes 65 | 66 | BOT: Instead of multiplying the elements directly, you can represent each element as $a_i \cdot 2^{x_i}$, where $x_i$ is the power to which you need to multiply the element. This way, you can calculate the minimum number of operations needed by adjusting $x_i$ based on the previous element's value. 67 | 68 | Would you like a more detailed hint on how to implement this approach? 69 | 70 | 71 | 72 | -------------------------------------------------------------------------------- /saved_chats/#7: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1981/F 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/129848 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: My first observation is that we can define contributing vertices as those that are < the MEX of their component. The answer is then contributing vertices + # of components, right? From here, we initialize every edge as its own node, and then merge edges with adjacent edges that form paths while the MEX of the merged path does not increase. I'm not sure if this solution is correct, and if it is I can't prove that is optimal. 20 | 21 | BOT: Your thoughts are not in the editorial. 22 | 23 | 24 | 25 | USER: Can you give me a small hint to help me proceed? 26 | 27 | BOT: Consider using dynamic programming (DP) to solve the problem. Define a DP state that helps you track the minimum result of the MEX of paths extending within subtrees. This will help you systematically build the solution. 28 | 29 | 30 | 31 | USER: What would such a DP look like? 32 | 33 | BOT: Define a DP state $ f_{u, i} $ which represents the minimum result of the MEX of a path that extends within the subtree rooted at vertex $ u $ and does not include the value $ i $. This will help you calculate the MEX values efficiently as you traverse the tree. 34 | 35 | 36 | 37 | USER: I see the transitions, but they are O(N^2) 38 | 39 | BOT: To optimize the transitions, consider that you only need to evaluate MEX values up to $ O\left(\frac{n}{\ln n}\right) $. This reduces the number of states you need to consider, making the solution more efficient. 40 | 41 | 42 | 43 | USER: Why is this true? 44 | 45 | BOT: The reason you only need to evaluate MEX values up to $ O\left(\frac{n}{\ln n}\right) $ is based on the distribution of values in the tree. As the tree grows, the number of distinct values that can appear on any path is limited by this bound. This makes it unnecessary to consider higher MEX values, as they will not be optimal. Would you like a more detailed explanation of this concept? 46 | 47 | 48 | 49 | -------------------------------------------------------------------------------- /saved_chats/#70002: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1942/problem/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | -------------------------------------------------------------------------------- /saved_chats/#71224: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1592/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/95583 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: i know the subarray length has to be even. i also think that i can iterate over the msb of the bitwise and, but i don't know how to check if the xor has a larger bit set. can i use a binary trie for this? 14 | 15 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 16 | 17 | 18 | 19 | USER: hint 20 | 21 | BOT: What are you current thoughts on the problem? 22 | 23 | 24 | 25 | USER: I know the subarray length has to be even. I also think that I can iterate over the MSB of the bitwise AND, but i don't know how to check if the XOR has a larger bit set. Can I use a binary trie for this? 26 | 27 | BOT: The idea of iterating over the MSB of the bitwise AND is on the right track. To check if the XOR has a larger bit set, you can use prefix arrays to keep track of the XOR values up to each index, considering only the important bits. 28 | 29 | For a hint: Focus on creating a prefix array where each entry represents the XOR of elements up to that index, considering only the bits more significant than the current bit you are examining. This will help you determine if the XOR of any subarray has a larger bit set. 30 | 31 | Would you like a more detailed explanation on how to construct and use this prefix array? 32 | 33 | 34 | 35 | USER: Can I use a binary trie for this? 36 | 37 | BOT: Using a binary trie is not mentioned as part of the solution approach in the editorial. Instead, the solution involves using prefix arrays to keep track of the XOR values up to each index, considering only the important bits. 38 | 39 | Would you like a more detailed hint on how to use these prefix arrays to solve the problem? 40 | 41 | 42 | 43 | -------------------------------------------------------------------------------- /saved_chats/#71632: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1521/A 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/90477 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: **Tutorial** 16 | There are $$$2$$$ cases: 17 | if $$$B = 1$$$, then the answer doesn't exist. Here we cannot get the nearly good numbers at all. 18 | Otherwise, we can construct the answer as $$$A + A \cdot B = A \cdot (B + 1)$$$. 19 | **Solution** 20 | Let me know if you have any questions. 21 | 22 | 23 | 24 | USER: test 25 | 26 | BOT: The problem requires finding three different positive integers $ x $, $ y $, and $ z $ such that exactly one of them is "good" (divisible by $ A \cdot B $), the other two are "nearly good" (divisible by $ A $), and they satisfy $ x + y = z $. The editorial outlines two cases: if $ B = 1 $, a solution is impossible because any number divisible by $ A $ would also be divisible by $ A \cdot B $, making it impossible to have "nearly good" numbers. For $ B > 1 $, the solution can be constructed as $ x = A $, $ y = A \cdot B $, and $ z = A \cdot (B + 1) $. Here, $ x $ and $ y $ are "nearly good" since they are divisible by $ A $, and $ y $ is also "good" since it is divisible by $ A \cdot B $. The sum $ x + y = A + A \cdot B = A \cdot (B + 1) = z $, ensuring $ z $ is the correct sum and is "nearly good". 27 | 28 | This approach ensures that the conditions are met: $ x $ and $ y $ are distinct, $ z $ is distinct, and only one number (in this case, $ y $) is "good" while the other two are "nearly good". The simplicity of the solution lies in leveraging the properties of multiplication and addition to construct the required numbers directly. 29 | 30 | 31 | 32 | -------------------------------------------------------------------------------- /saved_chats/#74733: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1980 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130135 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/contest/1980 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /saved_chats/#76032: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1980/problem/A 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/contest/1980 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | -------------------------------------------------------------------------------- /saved_chats/#8: -------------------------------------------------------------------------------- 1 | USER: https://usaco.org/index.php?page=viewproblem2&cpid=1421 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://usaco.org/current/data/sol_prob3_bronze_open24.html 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I figured out that using the first and last element, we can determine the sequence, so I can solve the problem by iterating over all pairs of first and last elements. But I don’t know how to optimize that further. 20 | 21 | BOT: You are on the right track by considering the first and last elements to determine the sequence. Here's a small hint to continue guiding you: 22 | 23 | Focus on the fact that the smallest element, 1, will never be removed and must be the last element written down. This means $ h_{N-1} $ must be 1. Use this observation to simplify your problem and discard the last element of $ h $. 24 | 25 | Would you like a more detailed hint on how to proceed from here? 26 | 27 | 28 | 29 | USER: But this still doesn't help me get the first and last element... 30 | 31 | BOT: To determine the first and last elements, observe that the numbers at the ends of $ p $ are precisely those two missing from $ h $. This means you can identify the first and last elements by finding the two numbers from $ 1 $ to $ N $ that are not present in $ h $. 32 | 33 | Would you like more detailed guidance on how to use this observation to reconstruct the permutation? 34 | 35 | 36 | 37 | -------------------------------------------------------------------------------- /saved_chats/#82434: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/blog/entry/130762 2 | 3 | BOT: There was something wrong with the link. Could you paste it again? 4 | 5 | 6 | 7 | USER: https://codeforces.com/contest/1978/problem/D 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | -------------------------------------------------------------------------------- /saved_chats/#86419: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/problemset/problem/1790/G 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/111948 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: a hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: I just need to find any token which has a path to node 1 with all nodes on the path being bonuses. And i can do this with dijkstra 20 | 21 | BOT: Your idea of using Dijkstra's algorithm to find a path to node 1 with all nodes on the path being bonuses aligns with the solution approach. You're thinking in the right direction! 22 | 23 | Hint: To proceed, consider how you can utilize the bonuses to extend your moves and how to handle the connected components of vertices with bonuses. This will help you determine the feasibility of reaching node 1. 24 | 25 | 26 | 27 | USER: explain the answer to me 28 | 29 | BOT: Sure, let's break down the solution step by step: 30 | 31 | 1. **Identify the Shortest Paths**: First, calculate the shortest paths from all vertices to the finish vertex (vertex 1) considering only the vertices that contain bonuses. This can be done using a modified version of Dijkstra's algorithm where you only traverse through vertices that have bonuses. 32 | 33 | 2. **Find the Closest Token**: Identify the token that is closest to the finish vertex. If there is no such token, it means you cannot reach the finish vertex, and thus you lose. 34 | 35 | 3. **Connected Components of Bonuses**: Find all connected components of vertices that contain bonuses. This helps in identifying which tokens can provide additional moves. 36 | 37 | 4. **Extra Moves Calculation**: For each connected component of bonuses, determine the tokens that are either in the component or adjacent to it. These tokens can provide extra moves. 38 | 39 | 5. **Component Size Analysis**: Analyze the size of each connected component: 40 | - If the size is 1, a token in an adjacent vertex provides 1 extra move. 41 | - If the size is greater than 1, tokens in the component or adjacent to it can move indefinitely, providing unlimited moves. 42 | 43 | 6. **Compare Moves and Distance**: Finally, compare the number of extra moves you have with the shortest distance to the finish vertex. If the extra moves are sufficient to reach the finish vertex, you win. 44 | 45 | Would you like more elaboration on any specific step? 46 | 47 | 48 | 49 | USER: thank you! 50 | 51 | BOT: You're welcome! If you have any more questions or need further clarification on any part of the solution, feel free to ask. Good luck with your problem-solving! 52 | 53 | 54 | 55 | -------------------------------------------------------------------------------- /saved_chats/#89801: -------------------------------------------------------------------------------- 1 | USER: https://usaco.org/index.php?page=viewproblem2&cpid=1110 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://usaco.org/current/data/sol_prob1_silver_feb21.html 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: Whenever there exists a cow horizontally or vertically adjacent to three other 16 | cows, Farmer Nhoj is forced to place a cow at the fourth spot. 17 | 18 | So this is essentially a flood fill problem; while there exists a location that 19 | should contain a cow but does not, add a cow at that location. 20 | My solution maintains a 2D boolean array of which locations currently contain 21 | cows, as well as a queue of locations in the pasture where a cow needs to be 22 | added. While the queue is nonempty, pop the top element $(x,y)$ of the queue and 23 | check whether a cow has already been added at $(x,y)$. If not, add the cow at 24 | $(x,y)$, and check whether any of the locations 25 | $(x,y),(x,y-1),(x,y+1),(x-1,y),(x+1,y)$ contain cows and are adjacent to exactly 26 | three cows. If so, add all such locations to the queue and repeat. 27 | Additional notes: 28 | The cows that will eventually be present on the pasture after the first $i$ 29 | cows are added is a superset of the cows that will eventually be present on the 30 | pasture after the first $i-1$ cows are added. 31 | This means that we don't need to reset the array between the addition of 32 | each cow.If all cells in $[0,1000]\times [0,1000]$ are initially occupied, Farmer 33 | Nhoj will need to add cows at all locations $(x,y)$ satisfying 34 | $x+y\ge 0,x+y\le 2000,x-y\le 1000,x-y\ge -1000$ (in other words, a diamond with 35 | vertices at $(-500,500), (500,-500), (1500,500),$ and $(500,1500)$). So if we 36 | increase the $x$ and $y$ cow locations by $500$, then all of these locations 37 | will lie in the range $[0,2000]\times [0,2000]$. 38 | In my solution, I use $1000$ instead of 39 | $500$. 40 | Time Complexity: $\mathcal{O}(N+(\text{grid size})^2)$. 41 | 42 | Here is an alternative solution by Danny Mittal that does not use a queue: 43 | 44 | 45 | Let me know if you have any questions. 46 | 47 | 48 | 49 | -------------------------------------------------------------------------------- /saved_chats/#90497: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1957/problem/E 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/128716 8 | 9 | BOT: There was something wrong with the link. Could you paste it again? 10 | 11 | 12 | 13 | USER: https://codeforces.com/blog/entry/128716 14 | 15 | BOT: There was something wrong with the link. Could you paste it again? 16 | 17 | 18 | 19 | USER: https://codeforces.com/blog/entry/128716 20 | 21 | BOT: There was something wrong with the link. Could you paste it again? 22 | 23 | 24 | 25 | USER: https://codeforces.com/blog/entry/128716 26 | 27 | BOT: There was something wrong with the link. Could you paste it again? 28 | 29 | 30 | 31 | USER: http://codeforces.com/blog/entry/128716 32 | 33 | BOT: There was something wrong with the link. Could you paste it again? 34 | 35 | 36 | 37 | FEEDBACK: it is not accepting the editorial. this is a pretty large flaw. 38 | 39 | 40 | 41 | -------------------------------------------------------------------------------- /saved_chats/#90715: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1978/problem/A 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/130527 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: hint 14 | 15 | BOT: What are you current thoughts on the problem? 16 | 17 | 18 | 19 | USER: please give some 20 | 21 | BOT: Sure, please share your current thoughts on solving the problem, and I can guide you from there. 22 | 23 | 24 | 25 | USER: i think first i have to sort, then pick such then pick such numbers which satisfy this question's answer 26 | 27 | BOT: You're on the right track by considering sorting. Here's a small hint to guide you further: Focus on the highest number in each pile. Since one pile will always include the book with the highest number of pages, think about how to maximize the sum of the highest numbers from both piles. Would you like a more detailed hint? 28 | 29 | 30 | 31 | USER: then i can sort upto second last index 32 | 33 | BOT: Yes, sorting up to the second last index is a good approach. Here's a more detailed hint: After sorting, you need to consider the book with the highest number of pages (which will be the last element after sorting) and pair it with the second highest number of pages (which will be the second last element after sorting). This will give you the maximum total number of pages Alice can read. Would you like further elaboration on this? 34 | 35 | 36 | 37 | USER: yes, please elaborate more on this 38 | 39 | BOT: Certainly! Here's a more detailed explanation: 40 | 41 | 1. **Sort the Books**: First, sort the list of books by the number of pages in ascending order. This will help you easily identify the books with the highest number of pages. 42 | 43 | 2. **Identify the Two Piles**: Since Alice will always read the book with the highest number of pages, this book will be in one pile. The next highest number of pages will be in the other pile to maximize the total number of pages read. 44 | 45 | 3. **Calculate the Maximum Pages**: After sorting, the book with the highest number of pages will be the last element in the sorted list. The book with the second highest number of pages will be the second last element in the sorted list. The maximum total number of pages Alice can read is the sum of these two values. 46 | 47 | For example, if the sorted list of pages is `[a1, a2, ..., an-1, an]`, the maximum total number of pages Alice can read is `an + an-1`. 48 | 49 | Would you like any further clarification or details on this approach? 50 | 51 | 52 | 53 | -------------------------------------------------------------------------------- /saved_chats/#94373: -------------------------------------------------------------------------------- 1 | USER: https://codeforces.com/contest/1942/problem/D 2 | 3 | BOT: Thanks! Now could you provide a link to the editorial. 4 | 5 | 6 | 7 | USER: https://codeforces.com/blog/entry/126942 8 | 9 | BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 10 | 11 | 12 | 13 | USER: editorial 14 | 15 | BOT: Problem Credits: sum Analysis: sum 16 | **Hint 1** 17 | We can solve this with dynamic programming. Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in order if you only paint cells $$$1,2,\ldots ,i$$$. 18 | **Hint 2** 19 | Transitioning boils down to finding the $$$k$$$ largest elements from $$$n$$$ lists. Try to do this in $$$\mathcal{O}((n + k) \log n)$$$ time. 20 | **Solution** 21 | Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in order if you only paint cells $$$1,2,\ldots ,i$$$. 22 | Let's define two lists as creating a new list that stores the $$$k$$$ highest values from the two lists. 23 | First, let's look at a slow method of transitioning. We iterate over the left endpoint $$$l$$$ such that $$$l \ldots i$$$ is a maximal painted interval. At each $$$l$$$, we merge $$$\texttt{dp}[l - 2]$$$, with $$$a_{l,i}$$$ added to each value, into $$$\texttt{dp}[i]$$$. We also merge $$$\texttt{dp}[i - 1]$$$ into $$$\texttt{dp}[i]$$$ to handle the case in which we do not paint cell $$$i$$$. If implemented well, transitioning takes $$$\mathcal{O}(nk)$$$ time leading to an $$$\mathcal{O}(n^2k)$$$ solution which is too slow. 24 | We can speed this up. This merging process boils down to finding the $$$k$$$ largest elements from $$$n$$$ lists in $$$\mathcal{O}((n + k) \log n)$$$ time. We can use a priority queue that stores ($$$\texttt{element}, \texttt{index}$$$) for each list. We can do the following $$$k$$$ times: add the top of the priority queue to our answer, advance its index, and update the priority queue accordingly. This allows us to transition in $$$\mathcal{O}((n + k) \log n)$$$ time which leads to an $$$\mathcal{O}((n^2 + nk) \log n)$$$ solution. 25 | Bonus: find an $$$\mathcal{O}(n^2 + k)$$$ solution. 26 | 27 | Let me know if you have any questions. 28 | 29 | 30 | 31 | USER: how do you know the priority queue thing is ideal when you also have to add the value from a? 32 | 33 | BOT: The priority queue approach is ideal because it efficiently merges the top elements from multiple lists, which is crucial for maintaining the highest beauty values. When transitioning, each list in the priority queue represents potential beauty values up to a certain cell. By storing elements as pairs (value, index), we can easily track and update the beauty values. For each cell $i$, we consider all possible left endpoints $l$ and add $a_{l,i}$ to the current beauty values from $\texttt{dp}[l-2]$. This ensures that the beauty values are updated correctly while maintaining the order of the highest values. 34 | 35 | The priority queue allows us to efficiently extract the top $k$ elements by always keeping the largest elements at the top. By repeatedly adding the top element to our result, advancing its index, and updating the priority queue, we ensure that the $k$ largest beauty values are always considered. This approach reduces the complexity of merging lists from $\mathcal{O}(nk)$ to $\mathcal{O}((n + k) \log n)$, making it much more efficient for large inputs. This efficiency is crucial for solving the problem within the given constraints. 36 | 37 | 38 | 39 | -------------------------------------------------------------------------------- /saved_chats/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/saved_chats/.DS_Store -------------------------------------------------------------------------------- /saved_chats/meta: -------------------------------------------------------------------------------- 1 | #13: 4 - Helpful 2 | #13: 4 - Helpful 3 | #19: 3 - Neutral 4 | #67753: 1 - Very Unhelpful 5 | #85873: 5 - Very Helpful 6 | #85873: 5 - Very Helpful 7 | #85873: 5 - Very Helpful 8 | #10265: 3 - Neutral 9 | #33526: 5 - Very Helpful 10 | #97244: 4 - Helpful 11 | #10008: 4 - Helpful 12 | #60851: 2 - Unhelpful 13 | #45407: 4 - Helpful 14 | #86419: 5 - Very Helpful 15 | #68745: Choose an Option 16 | #68745: Choose an Option 17 | #27080: 2 - Unhelpful 18 | #90497: 1 - Very Unhelpful 19 | #27974: 4 - Helpful 20 | -------------------------------------------------------------------------------- /utils/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/.DS_Store -------------------------------------------------------------------------------- /utils/__pycache__/chatbot.cpython-310.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/__pycache__/chatbot.cpython-310.pyc -------------------------------------------------------------------------------- /utils/__pycache__/chatbot.cpython-312.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/__pycache__/chatbot.cpython-312.pyc -------------------------------------------------------------------------------- /utils/__pycache__/prompts.cpython-310.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/__pycache__/prompts.cpython-310.pyc -------------------------------------------------------------------------------- /utils/__pycache__/prompts.cpython-312.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/__pycache__/prompts.cpython-312.pyc -------------------------------------------------------------------------------- /utils/chatbot.py: -------------------------------------------------------------------------------- 1 | import openai 2 | from openai import OpenAI 3 | import os 4 | openai.api_key = os.environ['OPENAI_KEY'] 5 | 6 | class Chat: 7 | 8 | def __init__(self, system_message, lim=0.005): 9 | self.messages = [ 10 | {"role": "system", "content" : system_message} 11 | ] 12 | self.left = lim 13 | self.client = OpenAI(api_key=API_KEY) 14 | 15 | def chat(self, query): 16 | if (self.left <= 0): 17 | return "You have reached the chat limit." 18 | if (len(query) > 20000): 19 | return "Chat input is too long" 20 | 21 | self.messages.append({"role" : "user", "content" : query}) 22 | 23 | # print(self.messages) 24 | 25 | response = self.client.chat.completions.create( 26 | model="gpt-4o-mini", 27 | messages=self.messages, 28 | temperature=0.3 29 | ) 30 | # del self.messages[-1] 31 | result = response.choices[0].message.content 32 | 33 | self.messages.append(response.choices[0].message) 34 | # print(response) 35 | # print([result]) 36 | 37 | result = result.replace('\\(', '$') 38 | result = result.replace('\\)', '$') 39 | # print([result]) 40 | 41 | self.left -= 0.6 * response.usage.completion_tokens / (10**6) + 0.15 * response.usage.prompt_tokens /(10**6) 42 | print(self.left) 43 | 44 | return (result) 45 | -------------------------------------------------------------------------------- /utils/scrapers/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/.DS_Store -------------------------------------------------------------------------------- /utils/scrapers/__pycache__/codeforces.cpython-310.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/__pycache__/codeforces.cpython-310.pyc -------------------------------------------------------------------------------- /utils/scrapers/__pycache__/codeforces.cpython-312.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/__pycache__/codeforces.cpython-312.pyc -------------------------------------------------------------------------------- /utils/scrapers/__pycache__/codeforces.cpython-37.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/__pycache__/codeforces.cpython-37.pyc -------------------------------------------------------------------------------- /utils/scrapers/__pycache__/usaco.cpython-310.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/__pycache__/usaco.cpython-310.pyc -------------------------------------------------------------------------------- /utils/scrapers/__pycache__/usaco.cpython-312.pyc: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/__pycache__/usaco.cpython-312.pyc -------------------------------------------------------------------------------- /utils/scrapers/edi_scraper.py: -------------------------------------------------------------------------------- 1 | from bs4 import BeautifulSoup 2 | import requests 3 | import json 4 | import sys 5 | import stat 6 | 7 | from selenium.webdriver import Chrome 8 | from selenium.webdriver.chrome.options import Options 9 | 10 | 11 | 12 | 13 | url = sys.argv[1] 14 | 15 | chrome_options = Options() 16 | chrome_options.add_argument("--headless") # Opens the browser up in background 17 | 18 | with Chrome(options=chrome_options) as browser: 19 | browser.get(url) 20 | response = browser.page_source 21 | 22 | 23 | # response = requests.get(url) 24 | # print(response) 25 | soup = BeautifulSoup(response, 'html.parser') 26 | 27 | if soup.text == "Just a moment...Enable JavaScript and cookies to continue": 28 | print("Bypassing anti-scrap protection...") 29 | scr = soup.findAll("script")[-1].string 30 | scr = scr[scr.index("var a=toNumbers"):].split(';') 31 | line = scr[0] 32 | abc = [] 33 | while "toNumbers" in line: 34 | i = line.index("toNumbers") 35 | line = line[i+11:] 36 | abc.append(line[:line.index('"')]) 37 | from Crypto.Cipher import AES 38 | def to_numbers(x): 39 | return bytes(int(x[i:i+2], 16) for i in range(0, len(x), 2)) 40 | key, iv, cipher = map(to_numbers, abc) 41 | aes = AES.new(key, AES.MODE_CBC, iv) 42 | rcpc = aes.decrypt(cipher).hex() 43 | print(f"RCPC = {rcpc}") 44 | url = scr[-2] 45 | url = url[url.index('"')+1:-1] 46 | r = requests.get(url, cookies={"RCPC": rcpc}) 47 | s = r.text 48 | soup = BeautifulSoup(s, "html.parser") 49 | 50 | soup = soup.find_all(class_='content')[0] # .find_all(class_ = 'spoiler') 51 | 52 | 53 | 54 | while soup.pre != None: 55 | soup.pre.decompose() 56 | 57 | import pickle 58 | 59 | 60 | # res=(soup.prettify().split(' a_n$$$, we have some observations: 8 | The leftmost element is non-decreasing. That is because if we want to remove the old leftmost element $$$a_1$$$, we need to pair it with $$$a_2 > a_1$$$, and that will result in the leftmost element increasing. 9 | Likewise, we can use the same argument to show that the rightmost element is non-increasing. 10 | Problem C author — Kuroni 11 | **Problem D** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1408D: -------------------------------------------------------------------------------- 1 | Let's define as $$$x$$$ our move to the right and as $$$y$$$ our move to the up. For all pairs $$$(i, j)$$$ of (robber, searchlight) at least one of this should be true: $$$x + a_i > c_j$$$, $$$y + b_i > d_j$$$. So if $$$x \leq c_j - a_i$$$ then $$$y \geq d_j - b_i + 1$$$. 2 | Let's make an array $$$r$$$ of length $$$C = 10^6$$$ and write on each position of $$$x$$$ the minimum value of $$$y$$$. For each $$$(i, j)$$$ we should make $$$r_x = max(r_x, d_j - b_i + 1)$$$ for all $$$x \leq c_j - a_i$$$. So we have $$$nm$$$ queries of $$$max=$$$ on prefix. We can do it using suffix maximums. After we will calculate all $$$a_x$$$ the answer is $$$\min\limits_{x}{(x + r_x)}$$$. 3 | Time complexity: $$$O(nm + C)$$$. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1492C: -------------------------------------------------------------------------------- 1 | For some subsequence of the string $$$s$$$, let $$$p_i$$$ denote the position of the character $$$t_i$$$ in the string $$$s$$$. For fixed $$$i$$$, we can find a subsequence that maximizes $$$p_{i + 1} - p_i$$$. 2 | Let $$$\textit{left}_{i}$$$ and $$$\textit{right}_{i}$$$ be the minimum and maximum possible value of $$$p_i$$$ among all valid $$$p$$$s. Now, it is easy to see that the maximum possible value of $$$p_{i+1} - p_{i}$$$ is equal to $$$\textit{right}_{i + 1} - \textit{left}_{i}$$$. 3 | To calculate $$$\textit{left}_i$$$ we just need to find the first element after $$$\textit{left}_{i - 1}$$$ which is equal to $$$t_i$$$. This can be done by a simple greedy or dp algorithm. $$$\textit{right}_i$$$ can be found in the same way. 4 | So, after finding $$$\textit{left}$$$ and $$$\textit{right}$$$, the answer is $$$\max_{i=1}^{i=n-1} \textit{right}_{i + 1} - \textit{left}_i$$$. 5 | Complexity: $$$O(n + m)$$$. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1521A: -------------------------------------------------------------------------------- 1 | **Tutorial** 2 | There are $$$2$$$ cases: 3 | if $$$B = 1$$$, then the answer doesn't exist. Here we cannot get the nearly good numbers at all. 4 | Otherwise, we can construct the answer as $$$A + A \cdot B = A \cdot (B + 1)$$$. 5 | **Solution** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1552D: -------------------------------------------------------------------------------- 1 | Suppose that a solution $$$b_1, \, \dots, \, b_n$$$ exists. For each $$$1 \le i \le n$$$, let $$$j_i, \, k_i$$$ be the indices such that $$$a_i = b_{j_i} - b_{k_i}$$$. 2 | Consider the directed graph on vertices $$$1, \, \dots, \, n$$$, with the $$$n$$$ edges $$$j_i \rightarrow k_i$$$. If we ignore, for a moment, the orientations, we are left with an undirected graph with $$$n$$$ vertices and $$$n$$$ edges, which must contain a cycle (possibly a loop). Let $$$m$$$ be the length of one such cycle, and let $$$v_1, \, \dots, v_m$$$ be its vertices. 3 | Now, of course $$$(b_{v_1} - b_{v_2}) + (b_{v_2} - b_{v_3}) + \cdots + (b_{v_m} - b_{v_1}) = 0$$$, since all the terms cancel out. Notice that, for each $$$i$$$, there exists $$$1 \le t_i \le n$$$ such that (indices are taken modulo $$$m$$$, so that $$$v_{m + 1} = v_1$$$) $$$$$$ b_{v_i} - b_{v_{i + 1}} = \begin{cases} a_{t_i} & \text{if there is an edge } v_i \rightarrow v_{i + 1}, \\ -a_{t_i} & \text{if there is an edge } v_i \leftarrow v_{i + 1}. \end{cases} $$$$$$ 4 | Thus, there must be a nonempty subset $$$\{t_1, \, \dots, \, t_m\} \subseteq \{1, \, \dots, \, n\}$$$ and a choice of signs $$$s_1, \, \dots, \, s_m$$$ ($$$s_i \in \{+1, \, -1\}$$$) such that $$$$$$\tag{$\star$} s_1a_{t_1} + \cdots + s_ma_{t_m} = 0. $$$$$$ 5 | Let us show that this condition is also sufficient for the existence of a valid sequence $$$b_1, \, \dots, \, b_n$$$. Suppose there exist $$$m$$$, $$$t_1, \, \dots, \, t_m$$$ and $$$s_1, \, \dots, \, s_m$$$ so that $$$(\star)$$$ holds. We construct $$$b_1, \, \dots, \, b_n$$$ as follows. Set $$$b_{t_1} = 0$$$. Then, inductively, for each $$$1 \le i < m$$$ set $$$b_{t_{i + 1}} = b_{t_i} - s_ia_{t_i}$$$. Finally, for $$$i \not\in \{t_1, \, \dots, \, t_m\}$$$, set $$$b_i = a_i$$$. It is easy to check that this construction works. 6 | The algorithm that iterates over all $$$3^n - 1$$$ choices of the subset and the signs (for each $$$i$$$ we decide whether $$$a_i$$$ is included in the subset and if it is included, whether its sign is positive or negative) and checks if for one of them $$$(\star)$$$ holds, is sufficient to solve the problem under the given constraints. 7 | Alternatively, one may treat the problem as a knapsack instance (the weights are $$$a_i$$$, but can be chosen with arbitrary sign, and we shall understand whether we can fill precisely a knapsack with $$$0$$$ capacity). With this approach, the complexity is $$$O(n^2\max|a_i|)$$$. 8 | Solve the problem with complexity $$$O(n3^{\frac{n}{2}})$$$. 9 | $$$O(n3^n)$$$. 10 | **Implementation** 11 | 123771071 12 | 13 | **E** 14 | **Hints for Solution 1** 15 | **Hint 1.** 16 | It is always convenient to choose $$$a_i, \, b_i$$$ so that the color $$$i$$$ does not appear in $$$a_i + 1, \, a_i + 2, \, \dots, \, b_i - 1$$$. So, for each color there are $$$k - 1$$$ "candidate intervals". 17 | **Hint 2.** 18 | $$$(k - 1)\left\lceil \frac{n}{k - 1} \right\rceil \ge n$$$. 19 | **Hint 3.** 20 | Choose the intervals in chunks of $$$\left\lceil \frac{n}{k - 1} \right\rceil$$$. The intervals in the first chunk should be strictly to the left of the intervals in the second chunk, which should be strictly to the left of the intervals of the third chunk, and so on. 21 | **Hint 4.** 22 | For the $$$t$$$-th chunk, only consider (among the remaining colors) the candidate intervals whose left endpoint is the $$$t$$$-th occurrence of its color. 23 | **Hints for Solution 2** 24 | **Hint 1.** 25 | Try to come up with a greedy algorithm. 26 | **Hint 2.** 27 | Order the candidate intervals (defined as above) by second endpoint. 28 | **Hint 3.** 29 | Scan the sorted intervals and always select one if you are allowed to. 30 | **Solution** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1592E: -------------------------------------------------------------------------------- 1 | Let $$$And(l, r)$$$ denotes the bitwise and of the elements in subarray $$$[L, R]$$$ in the array and $$$Xor(l, r)$$$ denotes the bitwise xor of the elements in subarray $$$[L, R]$$$ in the array. 2 | If subarray $$$[L, R]$$$ length is odd then it can't be a good subarray because $$$And(l, r)$$$ is a submask of $$$Xor(l, r)$$$ since every bit in $$$And(l, r)$$$ occurs odd times so it will be set in $$$Xor(l, r)$$$. 3 | If subarray $$$[L, r]$$$ length is even then every bit in $$$And(l, r)$$$ will be unset in $$$Xor(l, r)$$$ so we only care about the most significant bit in $$$And(l, r)$$$. 4 | Let's solve for every bit $$$k$$$, Let's call bit $$$m$$$ important bit if $$$m > k$$$. For every $$$r$$$ in the array, We need to find the minimum $$$l$$$ such that $$$[l, r]$$$ is even, $$$k$$$ is set in $$$And(l, r)$$$ and all the important bits are unset in $$$Xor(l, r)$$$. 5 | Since we only care about the important bits, We can make a prefix array where $$$pref_i$$$ is the prefix xor till the index $$$i$$$ considering only the important bits. 6 | So for every $$$r$$$, we need to find the minimum $$$l$$$ that satisfy these conditions: 7 | $$$1$$$ - $$$r-l+1$$$ is even. 8 | $$$2$$$ - $$$k$$$ is set in all elements in subarray $$$[l, r]$$$. 9 | $$$3$$$ - $$$pref_r$$$ xor $$$pref_{l-1}$$$ $$$= 0$$$. 10 | which can be solved easily in $$$O(NlogN)$$$. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1670E: -------------------------------------------------------------------------------- 1 | **Tutorial** 2 | Let's look at the minimum maximum value that we can get if we have an array of numbers from $$$[1,2^{(p+1)}-1]$$$ and we are trying to get any prefix xor, the answer will be $$$2^p$$$ because you can stop at the first integer that will have the bit $$$p$$$ so the answer will be $$$\geq 2^p$$$. 3 | We can apply the same concept here, for any arrangement we can start at the root and stop at the first node/edge that has the bit $$$p$$$ on. Let's try to find a construction that will make our answer always $$$2^p$$$. This is one of the valid ways. 4 | Select an arbitrary root. 5 | Put $$$2^p$$$ at the root. 6 | Create $$$2^p-1$$$ pairs from the remaining numbers of the form $$$(x,x+2^p)$$$ where $$$x < 2^p$$$ 7 | For every node we will do the following: If its parent has the bit $$$p$$$ in its value the node will take the value $$$x$$$ and the edge to the parent will take $$$x+2^p$$$. If its parent doesn't have the bit $$$p$$$ in its value the node will take the value $$$x+2^p$$$ and the edge to the parent will take $$$x$$$. Using this construction you will find that the xor value form the root will alternate between $$$0$$$ and $$$2^p$$$ and $$$x$$$ which is always $$$ \le 2^p$$$ . 8 | If its parent has the bit $$$p$$$ in its value the node will take the value $$$x$$$ and the edge to the parent will take $$$x+2^p$$$. 9 | If its parent doesn't have the bit $$$p$$$ in its value the node will take the value $$$x+2^p$$$ and the edge to the parent will take $$$x$$$. 10 | Using this construction you will find that the xor value form the root will alternate between $$$0$$$ and $$$2^p$$$ and $$$x$$$ which is always $$$ \le 2^p$$$ . 11 | **Solution** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1737B: -------------------------------------------------------------------------------- 1 | Author: low_ 2 | **Tutorial** 3 | We shift the perspective to see the pattern from $$$\lfloor \sqrt{x} \rfloor$$$, instead of $$$x$$$, to see what kind of patterns are needed so $$$x$$$ can be luxurious. 4 | Note that: $$${(a + 1)}^2 - 1 = a * (a + 2)$$$. 5 | Therefore, every half-closed segment $$$[a^2, (a + 1) ^ 2)$$$ contains $$$3$$$ luxurious number: $$$a^2$$$, $$$a * (a + 1)$$$ and $$$a * (a + 2)$$$. 6 | Also note that, since large numbers can cause inaccuracies in floating point computation, we should use binary search to find the floor-value of a square root, instead of using the sqrt function in any language. 7 | 8 | **Solution snippet (low_, C++)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1761E: -------------------------------------------------------------------------------- 1 | 2 | Try to figure out the conditions where a task can be solved with $$$1$$$ operation. Then $$$2$$$ operations, and then even more operations. 3 | 4 | The answer could be larger than $$$2$$$ only when the graph is made up of $$$2$$$ cliques, where you could only perform the operations on every vertex in the smaller clique to get the minimum number of operations. 5 | 6 | First of all, we need to check if the graph is already connected at the beginning. If so, the answer would be $$$0$$$. 7 | Otherwise, there will be more than $$$1$$$ connected component. If there exists a vertex that is the only vertex required to be operated to make the graph connected, we call such a vertex "feasible vertex". We may find out that a feasible vertex can only appear in a connected component that is not a clique. 8 | But actually, there will always be such a vertex in a non-clique component. To prove this, we may figure out the sufficient condition for being a feasible vertex first. 9 | The sufficient condition is that, if a vertex is not a cut vertex, and it is not adjacent to all other vertices in the connected component, then it must be a feasible vertex. We can prove that such a vertex always exists in a non-clique component. Here is the proof: 10 | Firstly, if there exist vertices that are adjacent to all other vertices in the component, we . Apparently, the remaining component would still be a non-clique component. Otherwise, the component could only be a clique from the beginning, which contradicts the premise. 11 | Apparently, the remaining component would still be a non-clique component. Otherwise, the component could only be a clique from the beginning, which contradicts the premise. 12 | Then, we will find a non-cut vertex in the remaining component, since that vertices in a graph couldn't be all cut vertices. The non-cut vertex we found is the vertex we are searching for. 13 | But implementing it directly (meaning using Tarjan's algorithm to find a non-cut vertex) might not be the easiest way to solve the problem. Actually, the vertex with the least degree in a connected component always satisfy the condition. We would like to leave the proof work of the alternative method to you. 14 | Now we have proven that, if there exists a connected component that is not a clique, then the answer would be at most $$$1$$$. What if all connected components are cliques? 15 | If there are exactly $$$2$$$ connected components, then apparently we will have to operate on all vertices in a connected component. So we'll choose the smaller connected component to operate, and the answer is exactly the size of it. 16 | Otherwise, we can arbitrarily choose two vertices from two different connected components and operate on them. The answer is $$$2$$$. 17 | Note that we also need to deal with isolated vertices (meaning vertices that are not adjacent to any other vertices) separately. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1790G: -------------------------------------------------------------------------------- 1 | Idea: senjougaharin 2 | **Tutorial** 3 | Let's calculate the shortest paths to the finish along the vertices containing bonuses. 4 | We will try to reach the finish line with the chip that is closest to it, and mark it. If there is none, we lose. 5 | Other chips will give her extra moves. 6 | Find all connected components from vertices containing bonuses. Then, for each component, we find all the tokens that are not selected, located at the vertex of this component, and at the vertices adjacent to at least one vertex of this component. 7 | Consider the size of the connectivity component. If it is equal to $$$1$$$, then the chip located in the neighboring vertex gives $$$1$$$ an additional move. Otherwise, the chip located at the top of the component or at the neighboring vertex, as well as the selected chip, will be able to move indefinitely in turn, which gives us a victory. Otherwise, you need to count the number of extra moves and compare it with the shortest distance to the finish line. 8 | **Solution** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1883E: -------------------------------------------------------------------------------- 1 | Let's come up with a naive solution — we could go from left to right and multiply the element at index $$$i$$$ by $$$2$$$ until it is greater than or equal to the previous element. But this solution will take a long time, as the numbers can become on the order of $$$2^n$$$ during the operations. 2 | Let's not explicitly multiply our numbers by $$$2$$$, but represent each element as the product $$$a_i \cdot 2^{x_i}$$$, where $$$x_i$$$ is the power to which we multiplied. Let's say we have calculated the value of $$$x_{i - 1}$$$, and now we want to calculate $$$x_i$$$. We have two cases: 3 | If $$$a_{i - 1} \leq a_i$$$: Then we can say that $$$x_i = x_{i - 1}$$$ and subtract $$$1$$$ until $$$x_i > 0$$$ and $$$a_{i - 1} \cdot 2 \leq a_i$$$. 4 | If $$$a_{i - 1} > a_i$$$: Then we can say that $$$x_i = x_{i - 1}$$$ and add $$$1$$$ until $$$a_i \cdot 2 < a_{i - 1}$$$. 5 | Note that we do not want to change the values of the array $$$a$$$ to avoid getting large numbers later, so we use additional variables for this. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1930B: -------------------------------------------------------------------------------- 1 | Idea: satyam343Editorial: satyam343 2 | **Hint 1** 3 | For integers $$$x$$$ ($$$\lfloor \frac{n}{2} \rfloor < x \leq n$$$), there does not exist integer $$$y$$$ ($$$y > x$$$) such that $$$y$$$ is divisible by $$$x$$$. 4 | **Solution** 5 | Consider the permutation $$$p$$$ such that $$$p=[1, n, 2, n - 1, \ldots \lceil \frac{n+1}{2} \rceil]$$$. It is valid. Why? 6 | **Proof** 7 | We have $$$\max(p_a, p_{a+1}) > \lfloor \frac{n}{2} \rfloor$$$ for all $$$1 \leq a < n - 1$$$. So we cannot ever have a pair of integers ($$$a,b$$$) such that: 8 | $$$1 \leq a < n - 1$$$ 9 | $$$1 \leq b < n$$$ 10 | $$$a \neq b$$$ 11 | $$$p_a$$$ divides $$$p_b$$$ and $$$p_{a+1}$$$ divides $$$p_{b+1}$$$ 12 | Now, we just need to check for $$$a = n - 1$$$. First of all, notice that $$$p_a$$$ does not divide $$$p_1$$$. 13 | There does not exist an integer $$$b$$$ ($$$2 \leq b < n - 1$$$) such that $$$p_{a+1}$$$ divides $$$p_{b+1}$$$ as $$$2 \cdot p_{a+1} \ge n$$$ and $$$p_{c+1} < n$$$ for all $$$c$$$ ($$$2 \leq c < n - 1$$$). 14 | Note that we covered all possible pairs of indices and did not find two indices $$$i$$$ and $$$j$$$ ($$$1 \leq i, j < n$$$; $$$i \neq j$$$) such that $$$p_i$$$ divides $$$p_j$$$ and $$$p_{i+1}$$$ divides $$$p_{j+1}$$$. 15 | 16 | video editorial by aryanc403 -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1942C1: -------------------------------------------------------------------------------- 1 | Problem Credits: cry Analysis: cry 2 | **Solution (Easy Version)** 3 | Ignoring $$$n$$$ for now, let's just focus on the $$$x$$$ chosen vertices. Sort the $$$x$$$ vertices and connect adjacent vertices to form its own smaller polygon. By drawing out some cases or if you're familiar with triangulation (video that proves by induction), you can form $$$x - 2$$$ triangles by drawing diagonals in a polygon with $$$x$$$ vertices. If you don't know it, one possible construction that always work is fixing one vertex $$$v$$$ and drawing diagonals to every other vertex not adjacent to $$$v$$$. Now, let's consider the original $$$n$$$-sided polygon. In additional to the aforementioned construction, to close the $$$x$$$-sided polygon up: for every non-adjacent vertex in the chosen vertices based on the bigger $$$n$$$-sided polygon, draw a diagonal between them. Through this construction, we can always guarantee $$$x - 2$$$ triangles. 4 | However, this doesn't account for all triangles, as some triangles can form sides with the bigger $$$n$$$-sided polygon. These triangles occur exactly when two adjacent vertex in $$$x$$$ have exactly one vertex not in $$$x$$$ between them but part of the bigger polygon. This is because one side is from the diagonals we drew, and the other two sides form from the $$$n$$$-sided polygon. 5 | Therefore, the answer is $$$x - 2$$$ + (number of adjacent chosen vertices $$$2$$$ apart). Note that the first chosen vertex is also adjacent to last chosen vertex in $$$x$$$. 6 | **Solution (Hard Version)** 7 | Read the solution to the easy version first. 8 | We can now reduce the problem to the following: For every vertex we choose, we can make one more triangle. For every chosen vertices two apart, we can make one more triangle. Let's focus on the latter condition. To maximize the effect of our $$$y$$$ vertices, we want to prioritize vertices $$$v$$$ that satisfy the following: the vertex is not included in $$$x$$$ and there is a chosen vertex two apart. Let's denote such $$$v$$$ as good. 9 | Let vertex $$$a$$$ and $$$b$$$ $$$(a < b)$$$ be adjacent chosen vertices in the $$$x$$$-sided polygon. There are $$$g = b - a - 1$$$ vertices for us to choose between them. There are two cases: if $$$g$$$ is odd, then we can choose $$$\lfloor \frac{g}{2} \rfloor$$$ vertices to form $$$\lfloor \frac{g}{2}\rfloor + 1$$$ extra triangles. This is because all of the vertices we choose will be good. if $$$g$$$ is even, then we can choose $$$\frac{g}{2}$$$ vertices to make $$$\frac{g}{2}$$$ extra triangles. This is because one of the vertices we choose will not be good. This shows that it is always optimal to process smaller and odd $$$g$$$ first. 10 | After processing these adjacent gaps, if we have any leftover vertices, we can just ignore them. This is because since we have maximized the number of good vertices, even though any further vertex we place will increase the answer by $$$1$$$, it will break two good vertices which will decrease the answer by $$$1$$$. 11 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1942D: -------------------------------------------------------------------------------- 1 | Problem Credits: sum Analysis: sum 2 | **Hint 1** 3 | We can solve this with dynamic programming. Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in order if you only paint cells $$$1,2,\ldots ,i$$$. 4 | **Hint 2** 5 | Transitioning boils down to finding the $$$k$$$ largest elements from $$$n$$$ lists. Try to do this in $$$\mathcal{O}((n + k) \log n)$$$ time. 6 | **Solution** 7 | Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in order if you only paint cells $$$1,2,\ldots ,i$$$. 8 | Let's define two lists as creating a new list that stores the $$$k$$$ highest values from the two lists. 9 | First, let's look at a slow method of transitioning. We iterate over the left endpoint $$$l$$$ such that $$$l \ldots i$$$ is a maximal painted interval. At each $$$l$$$, we merge $$$\texttt{dp}[l - 2]$$$, with $$$a_{l,i}$$$ added to each value, into $$$\texttt{dp}[i]$$$. We also merge $$$\texttt{dp}[i - 1]$$$ into $$$\texttt{dp}[i]$$$ to handle the case in which we do not paint cell $$$i$$$. If implemented well, transitioning takes $$$\mathcal{O}(nk)$$$ time leading to an $$$\mathcal{O}(n^2k)$$$ solution which is too slow. 10 | We can speed this up. This merging process boils down to finding the $$$k$$$ largest elements from $$$n$$$ lists in $$$\mathcal{O}((n + k) \log n)$$$ time. We can use a priority queue that stores ($$$\texttt{element}, \texttt{index}$$$) for each list. We can do the following $$$k$$$ times: add the top of the priority queue to our answer, advance its index, and update the priority queue accordingly. This allows us to transition in $$$\mathcal{O}((n + k) \log n)$$$ time which leads to an $$$\mathcal{O}((n^2 + nk) \log n)$$$ solution. 11 | Bonus: find an $$$\mathcal{O}(n^2 + k)$$$ solution. 12 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1949I: -------------------------------------------------------------------------------- 1 | To start, we build the graph that describes the tangency relation between the disks: each disk is represented by a node in the graph, and an edge connects two nodes if and only if the corresponding disks are tangent. Constructing this graph can be done in time $$$O(n^2)$$$ by checking all pairs of disks. In fact, it can be done in faster than quadratic time, but this is not necessary to solve the problem with the given constraints. 2 | Suppose that you change the radii of the disks by real numbers $$$\delta_1, \dots, \delta_n$$$. In order to maintain tangency, we must have $$$\delta_i + \delta_j = 0$$$ whenever the $$$i$$$-th and $$$j$$$-th disks are initially tangent. In fact, this is condition is also sufficient, provided that $$$\delta_1, \dots, \delta_n$$$ are small enough in absolute value (to avoid overlaps and to keep all the radii positive). 3 | Fix a connected component of the tangency graph, and fix any node $$$i$$$ in this component. The value $$$\delta_i$$$ determines the values of $$$\delta_j$$$ for all $$$j$$$ in the same connected component, and each $$$\delta_j$$$ has to be equal to either $$$\delta_i$$$ or $$$-\delta_i$$$. If the connected component is not bipartite, it contains a cycle of odd length, and such a cycle shows that $$$\delta_i = -\delta_i$$$ and so $$$\delta_i=0$$$. On the other hand, if the connected component is bipartite, then it is possible to assign any real value to $$$\delta_i$$$. Color the nodes of such a bipartite connected component black and white, so that every edge connects a white node and a black node; use white to color the node $$$i$$$. The sum of the radii in this connected component changes by $$$\delta_i \cdot k$$$, where $$$k$$$ is the difference between the number of white nodes and the number of black nodes. In order to strictly decrease the sum of the radii, we need a different number of black and white nodes ($$$k \neq 0$$$); if this happens, we can pick $$$\delta_i$$$ to be any sufficiently small real number such that $$$\delta_i \cdot k < 0$$$. 4 | To summarize, the answer is if and only if there is at least one connected component which is bipartite and has a different number of white and black nodes. This condition can be checked in time $$$O(n)$$$. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1957E: -------------------------------------------------------------------------------- 1 | SilverTongue1729 JadeReaper: SilverTongue1729 JadeReaper TheRaja 2 | **Hint 1** 3 | For what values of $$$j$$$ is $$$C(i,j) \bmod j = 0$$$? 4 | **Hint 2** 5 | For all other values of $$$j$$$, how can you precompute the result? 6 | **Hint 3** 7 | To precompute, switch around the loop order. 8 | **Solution** 9 | The number of distinct ways you can select $$$k$$$ distinct numbers from the set {$$$1, 2, \ldots, i$$$} and arrange them in a line is $$$i(i-1)\cdots(i-k+1)$$$, and since arranging in a circle introduces rotational symmetry we have to divide by $$$k$$$, so we have $$$C(i,k) = \frac{i(i-1)\cdots(i-k+1)}{k}$$$. 10 | Therefore $$$C(i,j) \bmod j = \frac{i(i-1)\cdots(i-j+1)}{j} \bmod j$$$. Now since the numerator is a product of $$$j$$$ consecutive integers, atleast one of them will be divisible by $$$j$$$. More precisely the exact integer which will be divisible by $$$j$$$ will be $$$ j \times \lfloor \frac{i}{j} \rfloor$$$. Hence we can simplify the fraction by removing the denominator and replacing the term $$$ j \times \lfloor \frac{i}{j} \rfloor$$$ with $$$\lfloor \frac{i}{j} \rfloor$$$ in the numerator. Each of the other $$$j-1$$$ integers in the numerator, after applying $$$\bmod j$$$ would cover all integers from $$$1$$$ to $$$j-1$$$. Hence 11 | $$$C(i,j) \bmod j = \frac{i(i-1)\cdots(i-j+1)}{j} \bmod j = \left( (j-1)! \times \left\lfloor \frac{i}{j} \right\rfloor \right) \bmod j$$$ 12 | Here we can notice that all proper factors of $$$j$$$ will occur in $$$(j-1)!$$$, so based on this we can tell that $$$C(i,j) \bmod j = 0$$$ for all composite numbers $$$j$$$ except $$$j=4$$$. 13 | We first can handle the case of $$$j$$$ being prime. Using Wilson's Theorem, we know that $$$(j-1)! \equiv -1 \bmod j$$$ when $$$j$$$ is prime. Hence 14 | $$$C(i,j) \bmod j = - \left\lfloor \frac{i}{j} \right\rfloor$$$ 15 | Now we can reverse the order of loops to sum over all primes, and to calculate the contribution of each prime we can maintain a update array called $$$delta$$$. 16 | To calculate the contribution for a single prime $$$p$$$, we know that for all $$$n$$$ from $$$kp$$$ to $$$(k + 1)p - 1$$$ (for all $$$k$$$ such that $$$kp < 1e6$$$) the contribution would be $$$-k$$$. So, in the $$$delta$$$ array, we increment index $$$kp$$$ with $$$-k \bmod p$$$ and decrement index $$$(k + 1)p$$$ with $$$-k \mod p$$$. Now, when we perform a prefix sum on this $$$delta$$$ array, we obtain the correct contributions from all primes. 17 | For the case of $$$j=4$$$, we just need to handle it as a prime. 18 | **Rate this problem** 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | Great Problem 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | Ok Problem 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | Bad Problem 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | Didn't solve 51 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1969C: -------------------------------------------------------------------------------- 1 | Idea: BledDest 2 | **Tutorial** 3 | The small values of $$$k$$$ leads to the idea that expected solution is dynamic programming. In fact, we can actually design a dynamic programming solution. 4 | Let $$$dp_{i, j}$$$ be the minimum sum, if we considered the first $$$i$$$ elements and already done $$$j$$$ operations. Note that, we can turn a segment of length $$$d+1$$$ into a minimum on it using $$$d$$$ operations. So the transitions can be done by iterating over the length of the next segment (denote it as $$$d$$$) and we can update $$$dp_{i + d + 1, j + d}$$$ with $$$dp_{i, j} + (d + 1) \cdot x$$$, where $$$x$$$ is the minimum among $$$a_i, a_{i + 1}, \dots, a_{i + d - 1}$$$ (that can be maintained in a single variable during iteration over $$$d$$$). 5 | There are $$$O(nk)$$$ states in the dynamic programming and $$$O(k)$$$ transitions from each of them, so the solution works in $$$O(nk^2)$$$. 6 | **Solution (Neon)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1978A: -------------------------------------------------------------------------------- 1 | Note that we will always be required to take a book with the number $$$n$$$. Thus, the answer can be all pairs of the form $$$(k, n)$$$, where $$$k < n$$$. We will find a pair with the maximum sum among these, and it will be the answer. 2 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1979A: -------------------------------------------------------------------------------- 1 | **Solution** 2 | Let $$$m$$$ be the maximum among the numbers $$$a_i, a_{i + 1},\ldots, a_j$$$. Notice that there always exists such $$$k$$$ that $$$i \le k < j$$$ and $$$a_k = m$$$ or $$$a_{k + 1} = m$$$. Therefore, we can assume that Bob always chooses the pair of numbers $$$p$$$ and $$$p + 1$$$ ($$$1 \le p < n$$$) as $$$i$$$ and $$$j$$$. 3 | Therefore you need to consider the maximums in pairs of adjacent elements and take the minimum among them. Let $$$min$$$ be the found minimum, then it is obvious that the answer is equal to $$$min - 1$$$. 4 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1979D: -------------------------------------------------------------------------------- 1 | **Solution** 2 | Let's consider the block of characters at the end. Notice that their quantity cannot decrease. Let $$$x$$$ be the number of identical characters at the end; there are three possible cases: 3 | $$$x = k$$$ — it is enough to find any block of length greater than $$$k$$$ and separate a block of length $$$k$$$ from it; 4 | $$$x > k$$$ — obviously, there is no solution; 5 | $$$x < k$$$ — find a block of length $$$k - x$$$ or $$$2k - x$$$ and separate a block of length $$$k - x$$$ from it. 6 | This solution works in $$$O(n)$$$, but it is not the only correct one. Your solution may differ significantly from the one proposed. 7 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1980E: -------------------------------------------------------------------------------- 1 | Idea: senjougaharin Prepared by: senjougaharin 2 | **Tutorial** 3 | For each element, you can calculate its positions in both matrices. You can see that the rearrangement of rows does not affect the column positions of the elements being rearranged. Similarly, column rearrangement does not affect row positions. Since the permutation of rows affects the entire rows, for all elements that have the same position row in the original matrix, the position row in the resulting matrix must also match. Similarly, the columns must match. 4 | In order to check the coincidence of rows and columns, let's count 4 arrays — the positions of rows and columns in the original and received matrices. Then you need to check that for all $$$x$$$ with the same row value in the original matrix, the row values in the resulting matrix are the same. Similarly, the values of the columns should be the same. 5 | **Solution** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1982D: -------------------------------------------------------------------------------- 1 | **Hint 1** 2 | How does the difference between heights change if we choose a specific submatrix? 3 | **Hint 2** 4 | It is worth reducing the problem to a mathematical construction. Maybe something from number theory can help? 5 | **Tutorial** 6 | First, let's calculate the current difference between the heights of different types of mountains, denoted as $$$D$$$. 7 | Then, for each submatrix of size $$$k \times k$$$, we will calculate how the difference in the sums of heights of different types of mountains changes. That is, for each submatrix, we will calculate the difference in the number of zeros and ones in it, let these values be $$$d_{i}$$$ and there will be $$$q = (n - k + 1) \cdot (m - k + 1)$$$. This can be done using two-dimensional prefix sums or a "sliding window". 8 | After that, formally we can write the following equation: 9 | $$$$$$ c_{1} \cdot d_{1} + c_{2} \cdot d_{2} + \dots + c_{q} \cdot d_{q} = D$$$$$$ 10 | , where $$$c_{1}$$$, $$$c_{2}$$$, ..., $$$c_{q}$$$ — are constants $$$c$$$ that Nikita chose for a specific submatrix ($$$0$$$ if we did not transform using this submatrix). This is very similar to a Diophantine equation, in fact, it is, but with $$$q$$$ unknowns. 11 | What can be said about the existence of a solution to this equation in integers? For this, it is sufficient that 12 | $$$$$$D \bmod gcd(d_{1}, d_{2}, \dots, d_{q}) = 0$$$$$$ 13 | It is worth mentioning that it is necessary to carefully handle the case when all $$$d_{i} = 0$$$, and also not to forget about the case when $$$D$$$ is initially equal to $$$0$$$. 14 | In total, the solution is in $$$O(n \cdot m + \log A)$$$ 15 | **Solution (FelixArg)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1982E: -------------------------------------------------------------------------------- 1 | **Hint** 2 | Try using the divide-and-conquer approach. 3 | **Tutorial** 4 | Let $$$f(n, k)$$$ be a function to solve the problem, which will return three values $$$(l, r, ans)$$$ such that: 5 | $$$l$$$ — for the first $$$l$$$ numbers (i.e. from $$$0$$$ to $$$l - 1$$$) $$$bit(x) \le k$$$ holds, while $$$bit(l) > k$$$ or $$$l \ge n$$$; 6 | $$$r$$$ — for the last $$$r$$$ numbers (i.e. from $$$n - r$$$ to $$$n - 1$$$) $$$bit(x) \le k$$$ holds, while $$$bit(n - r - 1) > k$$$ or $$$n - r - 1 < 0$$$; 7 | $$$ans$$$ — the answer for the pair $$$(n, k)$$$. 8 | Let's use the "divide and conquer" approach. Divide the segment from $$$0$$$ to $$$n$$$ into two segments: 9 | from $$$0$$$ to $$$2^{m} - 1$$$, where $$$m$$$ is the maximum such that $$$2^{m} < n$$$; 10 | from $$$2^{m}$$$ to $$$n - 1$$$. 11 | It can be noticed that instead of calculating the function on the segment from $$$2^{k}$$$ to $$$n - 1$$$, it is possible to calculate it from $$$0$$$ to $$$n - 1 - 2^{m}$$$, but with a smaller $$$k$$$. 12 | More formally, $$$f(n, k)$$$ can be recalculated through $$$f(2^{m}, k)$$$ and $$$f(n - 2^{m}, k - 1)$$$. 13 | Based on the pair of triples $$$(l_{1}, r_{1}, ans_{1})$$$, which is obtained after calculating $$$f(2^{m}, k)$$$ and $$$(l_{2}, r_{2}, ans_{2})$$$, which is obtained after calculating $$$f(n - 2^{m}, k - 1)$$$, it is quite trivial to recalculate the triple for $$$f(n, k)$$$ using the formula in $$$O(1)$$$. 14 | However, the solution still works in $$$O(n)$$$ time, as we will have a complete binary tree of recursive calls. 15 | To fix this, it is only necessary to precalculate all $$$f(2^{m}, k)$$$ for each possible $$$k$$$ and $$$m$$$. For $$$f(2^{m}, k)$$$, it can be noticed that it is recalculated through $$$f(2^{m - 1}, k)$$$ and $$$f(2^{m - 1}, k - 1)$$$, so all values can be calculated using dynamic programming. 16 | After this, in the recursive function, we always calculate $$$f(2^{m}, k)$$$ in $$$O(1)$$$, and for the next call, the first argument is halved, so now everything works in $$$O(\log n)$$$ for one test case. 17 | In total, the solution is in $$$O(\log n)$$$ for the test case. 18 | **Solution (FelixArg)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1982F: -------------------------------------------------------------------------------- 1 | **Hint** 2 | If you maintain a set of such indexes $$$i$$$ that $$$a_{i} pos_{max}$$$. How can we find exactly these $$$l$$$ and $$$r$$$? First, let's find $$$l$$$. To do this, we find the minimum on the segment $$$(pos_{min} - 1, pos_{max})$$$, this value will be somewhere in the prefix of the entire array, so we need to find the position in the prefix between which elements it should be inserted. But we also know that the prefix is sorted, so we can use binary search (as in insertion sort) and find the position for this value. This position will be our $$$l$$$. 9 | For $$$r$$$, everything is similar, we find the maximum on the segment $$$(pos_{min} - 1, pos_{max})$$$ and also find its position in the sorted suffix. 10 | In conclusion, to find the answer after changing the array, we need to find the maximum/minimum positions in $$$s$$$, the maximum/minimum on the segment, and perform $$$2$$$ binary searches on the prefix and suffix. All of this works in $$$O(\log n)$$$. 11 | Complexity: $$$O((n + q) \cdot \log n)$$$ 12 | It is also worth noting that some solutions with $$$O((n + q) \cdot \log^{2} n)$$$ can also fit within the time limit if they are very carefully written and use a non-recursive segment tree. 13 | **Solution (FelixArg)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1985G: -------------------------------------------------------------------------------- 1 | Problem Credits: cry Analysis: cry 2 | **Solution** 3 | To satisfy $$$D(k \cdot n) = k \cdot D(n)$$$, each digit $$$d$$$ in $$$n$$$ must become $$$k \cdot d$$$ after multiplying $$$n$$$ by $$$k$$$. In other words, none of $$$n$$$'s digits can carry over to the next digit upon multiplication. From this, we can deduce that each digit in $$$n$$$ must be less than or equal to $$$\lfloor \frac{9}{k} \rfloor$$$. Only thing left is to count all such numbers in the range of $$$10^l$$$ inclusive and $$$10^r$$$ exclusive. 4 | Every number below $$${10}^r$$$ has $$$r$$$ or less digits. For numbers with less than $$$r$$$ digits, let's pad the beginning with zeroes until it becomes a $$$r$$$ digit number (for example, if $$$r = 5$$$, then $$$69$$$ becomes $$$00069$$$). This allows us to consider numbers with less than $$$r$$$ digits the same way as numbers with exactly $$$r$$$ digits. For each digit, we have $$$\lfloor \frac{9}{k} \rfloor + 1$$$ choices (including zero), and there are $$$r$$$ digits, so the total number of numbers that satisfies the constraint below $$${10}^r$$$ is $$$(\lfloor \frac{9}{k} \rfloor + 1)^r$$$. 5 | To get the count of numbers in range, it suffices to subtract all valid numbers less than $$$10^l$$$. Therefore, the answer is $$$(\lfloor \frac{9}{k} \rfloor + 1)^r - (\lfloor \frac{9}{k} \rfloor + 1)^l$$$. To exponentiate fast, we can use modular exponentiation. 6 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1986D: -------------------------------------------------------------------------------- 1 | First, let's iterate through the position $$$i$$$, such that we do not place a mathematical sign between the $$$i$$$-th and $$$(i+1)$$$-th elements. 2 | Next, we have the following task — we have $$$n - 1$$$ numbers and we need to place a $$$+$$$ or $$$\times$$$ sign between each pair of neighboring numbers to minimize the result. There are three possible cases: 3 | If there is at least one $$$0$$$, the answer is $$$0$$$. We can simply place all signs as $$$\times$$$. 4 | If all numbers are $$$1$$$, the answer is $$$1$$$. We can simply place all signs as $$$\times$$$. 5 | Otherwise, the answer is the sum of numbers not equal to $$$1$$$. It is not advantageous to multiply numbers greater than one with each other, and all ones can simply be multiplied by any of the neighbors. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1987D: -------------------------------------------------------------------------------- 1 | **Hints** 2 | **Hint 1** 3 | Alice's optimal strategy is simple (and kind of fixed). 4 | **Hint 2** 5 | Alice always eats the least tasty cake that she can eat. 6 | **Hint 3** 7 | Convert the given array to it's count array $$$c_i$$$, for example, the array $$$[1, 5, 1, 2, 1, 5, 1]$$$ becomes $$$[4, 1, 2]$$$. 8 | **Hint 4** 9 | Consider a set of $$$k$$$ indices that Bob will choose to eat ($$$c_{i_1} + ... + c_{i_k}$$$ cakes in total). What are the necessary conditions for him to be able to do that? 10 | **Hint 5** 11 | For a set of $$$k$$$ indices $$$1 \le i_1 \le i_2 \le \ldots \le i_k \le |c|$$$, the condition $$$\sum_{j=1}^{p}{c_{i_j}} \le i_p - p$$$ must hold for all $$$1 \le p \le k$$$. 12 | **Tutorial** 13 | Let's consider Alice's strategy. Notice that if both players play optimally, Alice eating a cake with a tastiness value of $$$t$$$ is equivalent to her eating all remaining cakes with $$$a_i \le t$$$. Since it is better for her to have more cakes to choose from later on, she will choose the minimum possible $$$t$$$ on each turn. 14 | Now let's consider Bob's strategy. Suppose Bob ate at least one cake with a tastiness value of $$$t$$$. Then he has to have eaten all of them, because eating only some of them does not affect Alice's possible moves, resulting in wasted turns (he might be forced to waste moves at the end of the game when there are some leftover cakes). 15 | Let $$$A_1, \ldots, A_m$$$ be the sorted unique values of $$$a_1, \ldots, a_n$$$, and let $$$c_i$$$ be the number of occurrences of $$$A_i$$$ in $$$a$$$. Then the game is reduced to Alice finding the first $$$c_i > 0$$$ and assigning $$$c_i := 0$$$, and Bob choosing any $$$c_i > 0$$$ and decreasing it by $$$1$$$. 16 | Since Bob's optimal strategy is for each $$$c_i$$$ either not to touch it or make $$$c_i = 0$$$, we can model it as him selecting some set of indices $$$1 \le i_1 < \ldots < i_k \le m$$$ and zeroing out $$$c_{i_1}, c_{i_2}, \ldots, c_{i_k}$$$ in this order. To be able to zero out $$$c_{i_p}$$$ ($$$1 \le p \le k$$$), Alice must not have gotten to the value $$$c_{i_p}$$$. In $$$c_{i_1} + \ldots + c_{i_p}$$$ turns, Alice will zero out exactly that many values. Additionally, Bob would have zeroed out $$$p - 1$$$ values before index $$$i_p$$$. So, $$$c_{i_1} + \ldots + c_{i_p} + p - 1$$$ must be less than $$$i_p$$$. Transforming this a bit, we get that the condition $$$\sum_{j=1}^{p}{c_{i_j}} \le i_p - p$$$ must hold for all $$$1 \le p \le k$$$. 17 | Bob's objective to maximize the size of the set of incides $$$1 \le i_1 < \ldots < i_k \le m$$$. Let $$$dp[i][k] =$$$ the minimum possible $$$\sum_{j=1}^{k}{c_{i_j}}$$$ over all valid sets of indices $$$1 \le i_1 < \ldots < i_k \le i$$$. Initialize $$$dp[0][0] = 0$$$, and everything else to $$$+\infty$$$. The main idea is to grow valid sets of indices one index at a time. 18 | We will iterate over all $$$i$$$ from $$$1$$$ to $$$m$$$, and then for each $$$k$$$ from $$$0$$$ to $$$m$$$. The transitions are: 19 | $$$dp[i][k] = \min(dp[i][k], d[i - 1][k])$$$, which corresponds to not using the current index in the set. 20 | Let $$$s := dp[i - 1][k - 1] + c_i$$$. If $$$s \le i_k - k$$$, update $$$dp[i][k] = \min(dp[i][k], s)$$$, which is equivalent to adding the current index to the set. We only need to check the condition on the last index because it is already satisfied for all the previous indices. 21 | The answer to the problem is $$$m - y$$$, where $$$y$$$ is the largest value where $$$dp[m][y] \le +\infty$$$. 22 | Complexity: $$$\mathcal{O}(n^2)$$$ 23 | Note: it is possible to solve in $$$\mathcal{O}(n \log n)$$$ by making an additional observation. 24 | **Solution** 25 | **Feedback** 26 | Good problem 27 | 28 | 29 | 30 | 31 | 32 | Average problem 33 | 34 | 35 | 36 | 37 | 38 | Bad problem 39 | 40 | 41 | 42 | 43 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1989D: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/prescraped/codeforces/Editorials/1989D -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1989F: -------------------------------------------------------------------------------- 1 | Idea: BledDest 2 | **Tutorial** 3 | The main idea is as follows. Consider any constraint on a cell. If the cell must be red, then the last action on this cell must be coloring the row; otherwise, it should be the column. That is, if the operation is applied to both the row and the column of this cell, then there is a certain order of performing these operations. It is also worth noting that it does not make sense to apply the same operation twice, because the second application of the operation will recolor all the cells. 4 | The word "order" should immediately bring to mind thoughts of directed graphs and topological sorting. Let's try to build a graph of operation execution. Create a graph with $$$n + m$$$ vertices — one vertex for each row and column. If the cell must be red, then draw a directed edge from its column to its row. Otherwise, draw an edge from its row to its column. 5 | Now we would like to perform operations simply in the order of topological sorting. If this is possible (i.e., there are no cycles in the graph), then the answer is just $$$0$$$. This means that operations can always be applied one by one. It is also worth noting that we apply the operations that do not color any cells with constraints. Their cost is always $$$0$$$, so they do not change the answer. 6 | Otherwise, there are some strongly connected components in the graph with size greater than $$$1$$$. In these components, we will always have to perform some operations at the same time. Of course, operations can be applied to the entire component at once. The cost will then be equal to the square of the component's size. Doesn't sound optimal but that's one way to satisfy the constraints. 7 | Let's show that it is not possible to do it cheaper. Consider the last set of actions on this component. If they do not affect the entire component, then within the affected subset, there will be a vertex with an outgoing edge outside the subset, but inside the component (otherwise it would be two different components). This means that there must be at least one more operation performed on the vertex to which this edge leads. Therefore, the set of actions considered will not be the last one. 8 | Then the solution is as follows. After adding each constraint, add the previously described edges and find the condensation of the graph. The answer will be the sum of the squares of the sizes of strongly connected components with size greater than $$$1$$$. This solution works in $$$O(q \cdot (n + m + q))$$$. 9 | To optimize this solution, you can use the technique of incremental condensation. You can read about it in the blog by Radewoosh. Since DSU is used in it, you can maintain the sum of the squares of the component sizes on the fly. 10 | The blog describes a solution with a logarithmic time complexity, but the problem did not require an implementation faster than $$$log^2$$$. Solutions with a logarithmic memory complexity with a large constant might not pass (the author's solution consumes $$$O(q \log q)$$$ memory and fits into $$$50$$$ megabytes). 11 | **Solution (awoo)** -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/1994G: -------------------------------------------------------------------------------- 1 | **Hint 1** 2 | Think of the most stupid solution you can do 3 | **Hint 2** 4 | Do memorization and some cuts 5 | **Hint 3** 6 | Done! 7 | **Editorial** 8 | We will do recursion. Let's go from the lower bits to the significant bits. 9 | $$$rec(i, sum)$$$ means that the first $$$i$$$ bits are selected, the sum of them divided by $$$2^i$$$ equals $$$sum$$$, and the first $$$i$$$ bits match the ones we need. Let $$$cnt_i$$$ — the number of ones in the numbers in the $$$i$$$-th bit. Then we can go from state $$$(i, sum)$$$ to $$$(i + 1, \lfloor \frac{sum + cnt_i}{2} \rfloor)$$$ and $$$(i + 1, \lfloor \frac{sum + n - cnt_i}{2} \rfloor)$$$ if the lowest bit matches the $$$i$$$-th bit in $$$s$$$. There are at most $$$nk$$$ states in such a recursion (proof below). Thus, if we do the memoization, the solution will run $$$O(nk)$$$. 10 | Proof: 11 | $$$sum$$$ is always at most $$$n$$$. At the beginning of the recursion, $$$sum = 0$$$, so the inequality is satisfied. When we go to the next step $$$\frac{sum + cnt_i}{2}$$$, $$$\frac{sum + n - cnt_i}{2} \leq \frac{n + n}{2} = n$$$, which was required to prove. 12 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Editorials/992E: -------------------------------------------------------------------------------- 1 | This problem was inspired by idea which was offered by my unbelievable girlfriend :) 2 | 3 | In this problem we maintain two segment trees - with maximum and with sum. After every query we recalculate these trees in for a query. Now we only have to understand, how to answer for a query? Let's call a prefix of the array as in the case if we surely know that it doesn't contain a king-shaman. So either the first shaman is king and we are able to answer for the query, or we call the prefix with length 1 as . Then let's repeat the following operation: 4 | We call as sum in the good prefix now. We find it using the segment tree with sums. 5 | We find the leftmost element which may be king-shaman yet. We can realise that it's the leftmost element, which doesn't lie in the prefix (as there isn't king-shaman according the definition), which have a value at least . It can be done using segment tree with maximums, with descent. 6 | We check if this found shaman is king. If isn't, we can say that the good prefix finishes in this shaman now, because it was the leftmost shaman who might be king. 7 | Every operation works in . Let's notice, that one operation increases sum in the prefix at least by two times. So, after operations sum in the good prefix will become more than a maximal possible number in the array, so that we will be able to finish, because we will be sure that answer will be -1. Complexity of the solution is . 8 | 9 | Let be the prefix sums of . We're gonna introduce and reformulate the queries using these new terms. 10 | Imagine we wanna change the value at to . Let . Then will decrease by whereas will increase by . 11 | Imagine we wanna find the answer. Then it's sufficient to find any satisfying . 12 | Split into blocks of size . Each block will be comprised of pairs sorted by . At the same time we will maintain array responsible for lazy additions to blocks. How does this help? 13 | Let . The goal is to find the position of , decrease its value and increase values for all within this block. It can be done in a smart way in (actually, this can be represented as merging sorted vectors). You should also take care of the tail, i.e add to in time. 14 | We're asked to find such that . All blocks are sorted, thus we can simply apply binary search in overall. 15 | The optimal assignment is which results into total runtime. 16 | The huge advantage of this approach is its independency from constraints on (i.e non-negativity). Although we didn't succeed in squeezing this into TL :) 17 | 18 | Group the numbers according to their highest bits (with groups of the form and separately for zeros). Inside each groups (no more than of them) we maintain elements in increasing order of indexes. 19 | It's easy to see that each group contains no more than two candidates for answer (since their sum is guaranteed to be greater than any of the remaining ones). This observation leads to an easy solution in  — we iterate over groups and check prefix sums for these candidates. 20 | There's actually further space for optimizations. Let's maintain prefix sums for our candidates — this allows us to get rid of the extra log when quering the tree. Almost everything we need at this step is to somehow process additions and deletions — change the order inside two blocks and probably recalculate prefix sums. The only thing left is to stay aware of prefix sum changes for the remaining blocks. Luckily, they can be fixed in per block (if then the sum increases by and stays the same otherwise). 21 | The resulting comlexity is . 22 | Code of the solution I — 39423519 23 | Code of the solution II — 39418926. Try to optimize :) 24 | Thank you tfg for the idea and the code of the solution III. Very good job! 25 | Code of the solution III — 39392321 -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/prescraped/codeforces/Problems/.DS_Store -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1266A: -------------------------------------------------------------------------------- 1 | Bob is a competitive programmer. He wants to become red, and for that he needs a strict training regime. He went to the annual meeting of grandmasters and asked $$$n$$$ of them how much effort they needed to reach red."Oh, I just spent $$$x_i$$$ hours solving problems", said the $$$i$$$-th of them. Bob wants to train his math skills, so for each answer he wrote down the number of minutes ($$$60 \cdot x_i$$$), thanked the grandmasters and went home. Bob could write numbers with leading zeroes — for example, if some grandmaster answered that he had spent $$$2$$$ hours, Bob could write $$$000120$$$ instead of $$$120$$$.Alice wanted to tease Bob and so she took the numbers Bob wrote down, and for each of them she did one of the following independently: rearranged its digits, or wrote a random number. This way, Alice generated $$$n$$$ numbers, denoted $$$y_1$$$, ..., $$$y_n$$$.For each of the numbers, help Bob determine whether $$$y_i$$$ can be a permutation of a number divisible by $$$60$$$ (possibly with leading zeroes). -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1290D: -------------------------------------------------------------------------------- 1 | This is the hard version of the problem. You can find the easy version in the Div. 2 contest. Both versions only differ in the number of times you can ask your friend to taste coffee.This is an interactive problem.You're considering moving to another city, where one of your friends already lives. There are n cafés in this city, where n is a power of two. The i-th café produces a single variety of coffee ai. As you're a coffee-lover, before deciding to move or not, you want to know the number d of distinct varieties of coffees produced in this city.You don't know the values a1,…,an. Fortunately, your friend has a memory of size k, where k is a power of two.Once per day, you can ask him to taste a cup of coffee produced by the café c, and he will tell you if he tasted a similar coffee during the last k days.You can also ask him to take a medication that will reset his memory. He will forget all previous cups of coffee tasted. You can reset his memory at most 30 000 times.More formally, the memory of your friend is a queue S. Doing a query on café c will: Tell you if ac is in S; Add ac at the back of S; If |S|>k, pop the front element of S. Doing a reset request will pop all elements out of S.Your friend can taste at most 3n22k cups of coffee in total. Find the diversity d (number of distinct values in the array a).Note that asking your friend to reset his memory does not count towards the number of times you ask your friend to taste a cup of coffee.In some test cases the behavior of the interactor is adaptive. It means that the array a may be not fixed before the start of the interaction and may depend on your queries. It is guaranteed that at any moment of the interaction, there is at least one array a consistent with all the answers given so far. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1333C: -------------------------------------------------------------------------------- 1 | Eugene likes working with arrays. And today he needs your help in solving one challenging task.An array c is a subarray of an array b if c can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.Let's call a nonempty array good if for every nonempty subarray of this array, sum of the elements of this subarray is nonzero. For example, array [−1,2,−3] is good, as all arrays [−1], [−1,2], [−1,2,−3], [2], [2,−3], [−3] have nonzero sums of elements. However, array [−1,2,−1,−3] isn't good, as his subarray [−1,2,−1] has sum of elements equal to 0.Help Eugene to calculate the number of nonempty good subarrays of a given array a. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1375C: -------------------------------------------------------------------------------- 1 | You are given an array $$$a$$$ of length $$$n$$$, which initially is a permutation of numbers from $$$1$$$ to $$$n$$$. In one operation, you can choose an index $$$i$$$ ($$$1 \leq i < n$$$) such that $$$a_i < a_{i + 1}$$$, and remove either $$$a_i$$$ or $$$a_{i + 1}$$$ from the array (after the removal, the remaining parts are concatenated). For example, if you have the array $$$[1, 3, 2]$$$, you can choose $$$i = 1$$$ (since $$$a_1 = 1 < a_2 = 3$$$), then either remove $$$a_1$$$ which gives the new array $$$[3, 2]$$$, or remove $$$a_2$$$ which gives the new array $$$[1, 2]$$$.Is it possible to make the length of this array equal to $$$1$$$ with these operations? -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1408D: -------------------------------------------------------------------------------- 1 | There are n robbers at coordinates (a1,b1), (a2,b2), ..., (an,bn) and m searchlight at coordinates (c1,d1), (c2,d2), ..., (cm,dm). In one move you can move each robber to the right (increase ai of each robber by one) or move each robber up (increase bi of each robber by one). Note that you should either increase all ai or all bi, you can't increase ai for some points and bi for some other points.Searchlight j can see a robber i if ai≤cj and bi≤dj. A configuration of robbers is safe if no searchlight can see a robber (i.e. if there is no pair i,j such that searchlight j can see a robber i).What is the minimum number of moves you need to perform to reach a safe configuration? -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1474E: -------------------------------------------------------------------------------- 1 | Lunar rover finally reached planet X. After landing, he met an obstacle, that contains permutation p of length n. Scientists found out, that to overcome an obstacle, the robot should make p an identity permutation (make pi=i for all i).Unfortunately, scientists can't control the robot. Thus the only way to make p an identity permutation is applying the following operation to p multiple times: Select two indices i and j (i≠j), such that pj=i and swap the values of pi and pj. It takes robot (j−i)2 seconds to do this operation. Positions i and j are selected by the robot (scientists can't control it). He will apply this operation while p isn't an identity permutation. We can show that the robot will make no more than n operations regardless of the choice of i and j on each operation.Scientists asked you to find out the maximum possible time it will take the robot to finish making p an identity permutation (i. e. worst-case scenario), so they can decide whether they should construct a new lunar rover or just rest and wait. They won't believe you without proof, so you should build an example of p and robot's operations that maximizes the answer.For a better understanding of the statement, read the sample description. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1492C: -------------------------------------------------------------------------------- 1 | Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.A sequence p1,p2,…,pm, where 1≤p1al⊕al+1⊕al+2…⊕ar, where ⊕ denotes the bitwise XOR operation and & denotes the bitwise AND operation.Find the length of the longest good subarray of a, or determine that no such subarray exists. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1670E: -------------------------------------------------------------------------------- 1 | After the last regional contest, Hemose and his teammates finally qualified to the ICPC World Finals, so for this great achievement and his love of trees, he gave you this problem as the name of his team "Hemose 3al shagra" (Hemose on the tree).You are given a tree of n vertices where n is a power of 2. You have to give each node and edge an integer value in the range [1,2n−1] (inclusive), where all the values are distinct.After giving each node and edge a value, you should select some root for the tree such that the maximum cost of any simple path starting from the root and ending at any node or edge is minimized.The cost of the path between two nodes u and v or any node u and edge e is defined as the bitwise XOR of all the node's and edge's values between them, including the endpoints (note that in a tree there is only one simple path between two nodes or between a node and an edge). -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1737B: -------------------------------------------------------------------------------- 1 | While working at DTL, Ela is very aware of her physical and mental health. She started to practice various sports, such as Archery, Yoga, and Football.Since she started engaging in sports activities, Ela switches to trying a new sport on days she considers being "Luxury" days. She counts the days since she started these activities, in which the day she starts is numbered as day 1. A "Luxury" day is the day in which the number of this day is a luxurious number. An integer x is called a luxurious number if it is divisible by ⌊√x⌋.Here ⌊r⌋ denotes the "floor" of a real number r. In other words, it's the largest integer not greater than r.For example: 8, 56, 100 are luxurious numbers, since 8 is divisible by ⌊√8⌋=⌊2.8284⌋=2, 56 is divisible ⌊√56⌋=⌊7.4833⌋=7, and 100 is divisible by ⌊√100⌋=⌊10⌋=10, respectively. On the other hand 5, 40 are not, since 5 are not divisible by ⌊√5⌋=⌊2.2361⌋=2, and 40 are not divisible by ⌊√40⌋=⌊6.3246⌋=6.Being a friend of Ela, you want to engage in these fitness activities with her to keep her and yourself accompanied (and have fun together, of course). Between day l and day r, you want to know how many times she changes the activities. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1761E: -------------------------------------------------------------------------------- 1 | You are given a simple undirected graph consisting of n vertices. The graph doesn't contain self-loops, there is at most one edge between each pair of vertices. Your task is simple: make the graph connected.You can do the following operation any number of times (possibly zero): Choose a vertex u arbitrarily. For each vertex v satisfying v≠u in the graph individually, if v is adjacent to u, remove the edge between u and v, otherwise add an edge between u and v. Find the minimum number of operations required to make the graph connected. Also, find any sequence of operations with the minimum length that makes the graph connected. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1790G: -------------------------------------------------------------------------------- 1 | You are given an undirected connected graph, some vertices of which contain tokens and/or bonuses. Consider a game involving one player  — you.You can move tokens according to the following rules: At the beginning of the game, you can make exactly one turn: move any token to any adjacent vertex. If the movement of the token ended on the bonus, then you are allowed to make another turn with any other token. You can use different bonuses in any order. The same bonus can be used an unlimited number of times. Bonuses do not move during the game.There can be several tokens in one vertex at the same time, but initially there is no more than one token in each vertex.The vertex with number 1 is the finish vertex, and your task is to determine whether it is possible to hit it with any token by making turns with the tiles according to the rules described above. If a token is initially located at the vertex of 1, then the game is considered already won. The finish line is in black, the bonuses are in red, the chips are in grey. For example, for a given graph, you can reach the finish line with a chip from the 8th vertex by making the following sequence of turns: Move from the 8-th vertex to the 6-th. Move from the 7-th vertex to the 5-th. Move from the 6-th vertex to the 4-th. Move from the 5-th vertex to the 6-th. Move from the 4-th vertex to the 2-nd. Move from the 6-th vertex to the 4-th. Move from the 2-nd vertex to the 1-st vertex, which is the finish. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1883E: -------------------------------------------------------------------------------- 1 | You are given an array of integers a1,a2,…,an. You need to make it non-decreasing with the minimum number of operations. In one operation, you do the following: Choose an index 1≤i≤n, Set ai=ai⋅2.An array b1,b2,…,bn is non-decreasing if bi≤bi+1 for all 1≤i 1$$$, consisting of digits from $$$0$$$ to $$$9$$$. You must insert exactly $$$n - 2$$$ symbols $$$+$$$ (addition) or $$$\times$$$ (multiplication) into this string to form a valid arithmetic expression.In this problem, the symbols cannot be placed before the first or after the last character of the string $$$s$$$, and two symbols cannot be written consecutively. Also, note that the order of the digits in the string cannot be changed. Let's consider $$$s = 987009$$$: From this string, the following arithmetic expressions can be obtained: $$$9 \times 8 + 70 \times 0 + 9 = 81$$$, $$$98 \times 7 \times 0 + 0 \times 9 = 0$$$, $$$9 + 8 + 7 + 0 + 09 = 9 + 8 + 7 + 0 + 9 = 33$$$. Note that the number $$$09$$$ is considered valid and is simply transformed into $$$9$$$. From this string, the following arithmetic expressions cannot be obtained: $$$+9 \times 8 \times 70 + 09$$$ (symbols should only be placed between digits), $$$98 \times 70 + 0 + 9$$$ (since there are $$$6$$$ digits, there must be exactly $$$4$$$ symbols).The result of the arithmetic expression is calculated according to the rules of mathematics — first all multiplication operations are performed, then addition. You need to find the minimum result that can be obtained by inserting exactly $$$n - 2$$$ addition or multiplication symbols into the given string $$$s$$$. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1987D: -------------------------------------------------------------------------------- 1 | Alice and Bob are playing a game. Initially, there are n cakes, with the i-th cake having a tastiness value of ai.Alice and Bob take turns eating them, with Alice starting first: In her turn, Alice chooses and eats any remaining cake whose tastiness is strictly greater than the maximum tastiness of any of the cakes she's eaten before that. Note that on the first turn, she can choose any cake. In his turn, Bob chooses any remaining cake and eats it. The game ends when the current player can't eat a suitable cake. Let x be the number of cakes that Alice ate. Then, Alice wants to maximize x, while Bob wants to minimize x.Find out how many cakes Alice will eat if both players play optimally. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1987G1: -------------------------------------------------------------------------------- 1 | This is the easy version of the problem. The only difference between the two versions are the allowed characters in s. In the easy version, s only contains the character ?. You can make hacks only if both versions of the problem are solved.You are given a permutation p of length n. You are also given a string s of length n, consisting only of the character ?.For each i from 1 to n: Define li as the largest index jpi. If there is no such index, li:=i. Define ri as the smallest index j>i such that pj>pi. If there is no such index, ri:=i. Initially, you have an undirected graph with n vertices (numbered from 1 to n) and no edges. Then, for each i from 1 to n, add one edge to the graph: If si= L, add the edge (i,li) to the graph. If si= R, add the edge (i,ri) to the graph. If si= ?, either add the edge (i,li) or the edge (i,ri) to the graph at your choice. Find the maximum possible diameter∗ over all connected graphs that you can form. Output −1 if it is not possible to form any connected graphs.∗ Let d(s,t) denote the smallest number of edges on any path between s and t.The diameter of the graph is defined as the maximum value of d(s,t) over all pairs of vertices s and t. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1987H: -------------------------------------------------------------------------------- 1 | This temple only magnifies the mountain's power.BadelineThis is an interactive problem.You are given two positive integers n and m (n≤m).The jury has hidden from you a rectangular matrix a with n rows and m columns, where ai,j∈{−1,0,1} for all 1≤i≤n and 1≤j≤m. The jury has also selected a cell (i0,j0). Your goal is to find (i0,j0).In one query, you give a cell (i,j), then the jury will reply with an integer. If (i,j)=(i0,j0), the jury will reply with 0. Else, let S be the sum of ax,y over all x and y such that min(i,i0)≤x≤max(i,i0) and min(j,j0)≤y≤max(j,j0). Then, the jury will reply with |i−i0|+|j−j0|+|S|. Find (i0,j0) by making at most n+225 queries.Note: the grader is not adaptive: a and (i0,j0) are fixed before any queries are made. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1989B: -------------------------------------------------------------------------------- 1 | You are given two strings a and b, both consisting of lowercase Latin letters. A subsequence of a string is a string which can be obtained by removing several (possibly zero) characters from the original string. A substring of a string is a contiguous subsequence of that string.For example, consider the string abac: a, b, c, ab, aa, ac, ba, bc, aba, abc, aac, bac and abac are its subsequences; a, b, c, ab, ba, ac, aba, bac and abac are its substrings. Your task is to calculate the minimum possible length of the string that contains a as a substring and b as a subsequence. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1989D: -------------------------------------------------------------------------------- 1 | You are playing a famous computer game (that just works) where you have various skills you can level up. Today, you focused on the "Smithing" skill. Your tactic is obvious: forging weapons from ingots and then melting them back to return the materials partially. For simplicity, every time you create an item, you get 1 experience point, and every time you melt an item, you also get 1 experience point.There are n classes of weapons you can forge and m types of metal ingots.You can create one weapon of the i-th class, spending ai ingots of metal of the same type. Melting a weapon of the i-th class (which you crafted earlier) returns you bi ingots of the type of metal it was made of.You have cj metal ingots of the j-th type, and you know that you can craft a weapon of any class from any metal type. Each combination of a weapon class and a metal type can be used any number of times.What is the maximum total amount of experience you can earn by crafting and melting weapons? -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1989F: -------------------------------------------------------------------------------- 1 | You are given a matrix, consisting of n rows and m columns.You can perform two types of actions on it: paint the entire column in blue; paint the entire row in red. Note that you cannot choose which color to paint the row or column.In one second, you can perform either one action or multiple actions at the same time. If you perform one action, it will be free. If you perform k>1 actions at the same time, it will cost k2 coins. When multiple actions are performed at the same time, for each cell affected by actions of both types, the color can be chosen independently.You are asked to process q queries. Before each query, all cells become colorless. Initially, there are no restrictions on the color of any cells. In the i-th query, a restriction of the following form is added: xi yi ci — the cell in row xi in column yi should be painted in color ci. Thus, after i queries, there are i restrictions on the required colors of the matrix cells. After each query, output the minimum cost of painting the matrix according to the restrictions. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1994G: -------------------------------------------------------------------------------- 1 | After winning another Bed Wars game, Masha and Olya wanted to relax and decided to play a new game. Masha gives Olya an array a of length n and a number s. Now Olya's task is to find a non-negative number x such that n∑i=1ai⊕x=s. But she is very tired after a tight round, so please help her with this.But this task seemed too simple to them, so they decided to make the numbers larger (up to 2k) and provide you with their binary representation. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/1997F: -------------------------------------------------------------------------------- 1 | You have n chips, and you are going to place all of them in one of x points, numbered from 1 to x. There can be multiple chips in each point.After placing the chips, you can perform the following four operations (in any order, any number of times): choose a chip in point i≥3, remove it and place two chips: one in i−1, one in i−2; choose two chips in adjacent points i and i+1, remove them and place a new chip in i+2; choose a chip in point 1 and move it to 2; choose a chip in point 2 and move it to 1. Note that the coordinates of the chips you place during the operations cannot be less than 1, but can be greater than x.Denote the cost of chip placement as the minimum number of chips which can be present on the line after you perform these operations, starting from the placement you've chosen.For example, the cost of placing two chips in points 3 and one chip in point 5 is 2, because you can reduce the number of chips to 2 as follows: choose a chip in point 3, remove it, place a chip in 1 and another chip in 2; choose the chips in points 2 and 3, remove them and place a chip in 4; choose the chips in points 4 and 5, remove them and place a chip in 6. You are given three integers n, x and m. Calculate the number of placements of exactly n chips in points from 1 to x having cost equal to m, and print it modulo 998244353. Two placements are considered different if the number of chips in some point differs between these placements. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/codeforces/Problems/992E: -------------------------------------------------------------------------------- 1 | Nastya likes reading and even spends whole days in a library sometimes. Today she found a chronicle of Byteland in the library, and it stated that there lived shamans long time ago. It is known that at every moment there was exactly one shaman in Byteland, and there were n shamans in total enumerated with integers from 1 to n in the order they lived. Also, each shaman had a magic power which can now be expressed as an integer.The chronicle includes a list of powers of the n shamans. Also, some shamans can be king-shamans, if they gathered all the power of their predecessors, i.e. their power is exactly the sum of powers of all previous shamans. Nastya is interested in whether there was at least one king-shaman in Byteland.Unfortunately many of the powers are unreadable in the list, so Nastya is doing the following: Initially she supposes some power for each shaman. After that she changes the power of some shaman q times (the shamans can differ) and after that wants to check if there is at least one king-shaman in the list. If yes, she wants to know the index of any king-shaman. Unfortunately the list is too large and Nastya wants you to help her. -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/prescraped/usaco/.DS_Store -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Editorials/.DS_Store: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/prescraped/usaco/Editorials/.DS_Store -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Editorials/index.php?page=viewproblem2&cpid=1381: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Rithwik-G/competitive-programming-bot/93fa496a49836fc0d6e95ccfdf91f1f03e4c0453/utils/scrapers/prescraped/usaco/Editorials/index.php?page=viewproblem2&cpid=1381 -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Editorials/sol_prob1_silver_feb21.html: -------------------------------------------------------------------------------- 1 | Whenever there exists a cow horizontally or vertically adjacent to three other 2 | cows, Farmer Nhoj is forced to place a cow at the fourth spot. 3 | 4 | So this is essentially a flood fill problem; while there exists a location that 5 | should contain a cow but does not, add a cow at that location. 6 | My solution maintains a 2D boolean array of which locations currently contain 7 | cows, as well as a queue of locations in the pasture where a cow needs to be 8 | added. While the queue is nonempty, pop the top element $(x,y)$ of the queue and 9 | check whether a cow has already been added at $(x,y)$. If not, add the cow at 10 | $(x,y)$, and check whether any of the locations 11 | $(x,y),(x,y-1),(x,y+1),(x-1,y),(x+1,y)$ contain cows and are adjacent to exactly 12 | three cows. If so, add all such locations to the queue and repeat. 13 | Additional notes: 14 | The cows that will eventually be present on the pasture after the first $i$ 15 | cows are added is a superset of the cows that will eventually be present on the 16 | pasture after the first $i-1$ cows are added. 17 | This means that we don't need to reset the array between the addition of 18 | each cow.If all cells in $[0,1000]\times [0,1000]$ are initially occupied, Farmer 19 | Nhoj will need to add cows at all locations $(x,y)$ satisfying 20 | $x+y\ge 0,x+y\le 2000,x-y\le 1000,x-y\ge -1000$ (in other words, a diamond with 21 | vertices at $(-500,500), (500,-500), (1500,500),$ and $(500,1500)$). So if we 22 | increase the $x$ and $y$ cow locations by $500$, then all of these locations 23 | will lie in the range $[0,2000]\times [0,2000]$. 24 | In my solution, I use $1000$ instead of 25 | $500$. 26 | Time Complexity: $\mathcal{O}(N+(\text{grid size})^2)$. 27 | 28 | Here is an alternative solution by Danny Mittal that does not use a queue: 29 | 30 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Editorials/sol_prob2_platinum_jan24.html: -------------------------------------------------------------------------------- 1 | Note: let $\texttt{sum}(l, r)$ denote $s_l + s_{l+1} + \dots + s_{r-1} + s_r$. 2 | Subtask 1 ($N\leq 8$) 3 | It suffices to brute force all $(N-1)!$ permutations of merges. 4 | Subtask 2 ($N\leq 100$) 5 | Let's try to calculate the answer for a given cell $c$. At any step, $c$ must be 6 | the label of a cell that represents the sum of a range $[l, r]$ in the original 7 | array. Let $\texttt{dp}[l][r]$ represent the probability that $c$ becomes the 8 | final label if we only consider the cells in the range $[l, r]$. For 9 | transitions, we can consider enumerating the last merge while making sure that 10 | the $c$ remains as the label. Also, note that each of the $r-l$ possible merges 11 | has an equal probability of being the last merge. 12 | The base case is $\texttt{dp}[c][c] = 1$. Otherwise, we have the following: 13 | $$ 14 | \texttt{dp}[l][r] = 15 | \sum_{\substack{m\in (l, c] \\ \texttt{sum}(l, m-1)\leq \texttt{sum}(m, r)}}{\frac{1}{r-l}\cdot\texttt{dp}[m][r]} + 16 | \sum_{\substack{m\in [c, r] \\ \texttt{sum}(l, m) > \texttt{sum}(m+1, r)}}{\frac{1}{r-l}\cdot\texttt{dp}[l][m]} 17 | 18 | $$ 19 | This runs in $\mathcal O(N^3)$ for a single cell $c$, resulting in a 20 | $\mathcal O(N^4)$ solution. 21 | My code: 22 | 23 | Subtask 3 ($N\leq 500$) 24 | It is possible to optimize the solution from subtask 2 to $\mathcal O(N^3)$ 25 | using prefix sums and two pointers (described more in detail in subtask 4's 26 | notes). 27 | However, I will discuss another approach that is closer to the full solution. 28 | Notice that recalculating the DP for each $c$ can be suboptimal in many ways; 29 | for example, regardless of which label a cell $(l, r)$ contains, it has the same 30 | contribution to $\texttt{dp}[1][N]$. Therefore, let's consider calculating the 31 | DP backwards. 32 | Let $\texttt{dp}[l][r]$ represent the probability that the label of the cell 33 | representing the range $[l, r]$ becomes the label of the final cell after all 34 | merges. The base case is $\texttt{dp}[1][N] = 1$. Otherwise, transitions are 35 | similar to the ones described in the subtask 2 solution: 36 | $$ 37 | \texttt{dp}[l][r] = 38 | \sum_{\substack{p\in [1, l) \\ \texttt{sum}(l, r)\geq \texttt{sum}(p, l-1)}}{\frac{1}{r-p}\cdot \texttt{dp}[p][r]} + 39 | \sum_{\substack{p\in (r, N] \\ \texttt{sum}(l, r) > \texttt{sum}(r+1, p)}}{\frac{1}{p-l}\cdot \texttt{dp}[l][p]} 40 | 41 | $$ 42 | The answer for a cell $c$ is $\texttt{dp}[c][c]$. This runs in 43 | $\mathcal O(N^3)$. 44 | My code: 45 | 46 | Subtask 4 $(N\leq 5000$) 47 | Let's try to accelerate the solution from subtask 3. For example, say the last 48 | merge is leftwards. We query the sum of $\frac{\texttt{dp}[p][r]}{r-p}$ over a fixed $r$ and 49 | a range of $p$. The range of $p$ is determined by the leftmost position such 50 | that $\texttt{sum}(l, r)\geq \texttt{sum}(p, l-1)$ holds. Let this bound be 51 | denoted as $\texttt{pL}[l][r]$. It can be proven that 52 | $\texttt{pL}[l][r]\leq \texttt{pL}[l+1][r]$, which allows us to compute all 53 | $\texttt{pL}[l][r]$ in $\mathcal O(N^2)$ time using two pointers. Next, to 54 | calculate the sum, we need a data-structure that can support adding 55 | $\frac{\texttt{dp}[l][r]}{r-l}$ and querying the sum over 56 | $p\in [\texttt{pL}[l][r], l)$. Since we enumerate ranges in decreasing order of 57 | size, for a fixed $r$, $\frac{\texttt{dp}[l][r]}{r-l}$ is added in increasing order of $l$. 58 | This allows us to maintain prefix sums and answer range sum queries in constant 59 | time. 60 | This runs in $\mathcal O(N^2)$. My code: 61 | 62 | The code above uses around $5N^2$ memory. It's possible to optimize this to 63 | $N^2$ by just storing the DP array while maintaining $\texttt{pL}[l][r]$, 64 | $\texttt{pR}[l][r]$, and the range sums in 1D arrays. 65 | Benjamin Qi's code: 66 | 67 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Editorials/sol_prob3_bronze_open24.html: -------------------------------------------------------------------------------- 1 | Subtask 1: Try all $N!=N\cdot (N-1)\cdots 1$ possible $p$ in 2 | lexicographical order, and see if they can produce $h$. In Python, we can use 3 | itertools.permutations 4 | to do this easily. 5 | Ben's Python code: 6 | 7 | Subtask 2: If we are given the values of $p_1$ and $p_N$, we can uniquely 8 | determine the rest of $p$ by going through $h$ one element at a time. In 9 | particular, when $N>2$, 10 | If $p_1>p_N$, then $p_2=h_1$. It remains to determine the elements between 11 | $p_2$ and $p_N$.Else, $p_{N-1}=h_1$. It remains to determine the elements between $p_1$ and 12 | $p_{N-1}$ 13 | and then we can process $h_2,h_3,\dots, h_{N-1}$ similarly. At the end, we 14 | should check if $p$ forms a permutation or not and actually produces $h$; if so, 15 | we update our answer. The time complexity is $O(N^3)$ per test. 16 | Ben's Python code: 17 | 18 | Subtask 3: Consider the process of producing $h$ from $p$. Note that the 19 | smallest element of $p$, $1$, will never be removed. So when two elements remain 20 | in $p$, one must be $1$, and $1$ will be the last element written down. Thus, we 21 | must have $h_{N-1}=1$, or else we can output $-1$ and exit. For the remainder of 22 | the problem, we will not use the last element of $h$, so we discard the last 23 | element of $h$ and assume that $h$ has length $N-2$ going forward. 24 | If the answer is not $-1$, the lexicographically minimum $p$ must satisfy 25 | $p_10$ for all $l < |seq|$.$N+\sum_{t=0}^{|seq|-1}\texttt{changes}[t][seq[t]] + \sum_{t=|seq|}^{l-1}\max(\texttt{changes}[t][0], \texttt{changes}[t][1])>0$ 48 | for all $|seq|\le l\le M$. 49 | Suppose that the first condition already holds; then the second condition holds 50 | if and only if the minimum prefix sum (including the empty prefix) of the array 51 | 52 | is greater than some threshold. Let $\texttt{min_psum}[t]$ denote the minimum 53 | prefix sum when $|seq| = t$; then we have 54 | $\texttt{min_psum}[t]=\min(0, \max(\texttt{changes}[t]) + \texttt{min_psum}[t+1])$ 55 | for $t c_b \\ 19 | b & c_a < c_b \\ 20 | \max(a,b) & c_a = c_b 21 | \end{cases}.$ 22 | 23 | For each label $i$ in the range $1\dots N$, the probability that the final cell 24 | has label $i$ can be expressed in the form $\frac{a_i}{b_i}$ where 25 | $b_i\not\equiv 0\pmod{10^9+7}$. Output $a_ib_i^{-1}\pmod{10^9+7}$. 26 | 27 | INPUT FORMAT (input arrives from the terminal / stdin): 28 | The first line contains $N$. 29 | 30 | The next line contains $s_1,s_2,\dots, s_N$. 31 | 32 | OUTPUT FORMAT (print output to the terminal / stdout): 33 | The probability of the final cell having label $i$ modulo $10^9+7$ for each $i$ in $1\dots N$ on 34 | separate lines. 35 | 36 | SCORING: 37 | Input 3: $N\le 8$Inputs 4-8: $N\le 100$Inputs 9-14: $N\le 500$Inputs 15-22: No additional constraints. 38 | 39 | 40 | Problem credits: Benjamin Qi 41 | 42 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Problems/index.php?page=viewproblem2&cpid=1400: -------------------------------------------------------------------------------- 1 | 2 | 3 | Bessie and Elsie are playing a game of Moorbles. The game works as follows: 4 | Bessie and Elsie each start out with some amount of marbles. Bessie holds out 5 | $A$ of her marbles in her hoof and Elsie guesses if $A$ is Even or Odd. If Elsie 6 | is correct, she wins the $A$ marbles from Bessie and if she guesses incorrectly, 7 | she loses $A$ of her marbles to Bessie (if Elsie has less than $A$ marbles, she 8 | loses all her marbles). A player loses when they lose all of their marbles. 9 | 10 | After some amount of turns in the game, Elsie has $N$ $(1 \leq N \leq 10^9)$ 11 | marbles. She thinks it is hard to win, but she is playing to not lose. After 12 | being around Bessie enough, Elsie has a good read on Bessie's habits and 13 | recognizes that on turn $i$, there are only $K$ $(1 \leq K \leq 4)$ different 14 | amounts of marbles that Bessie may put out. There are only $M$ 15 | $(1 \leq M \leq 3 \cdot 10^5)$ turns before Bessie gets bored and stops playing. 16 | Can you identify a lexicographically minimum turn sequence such that Elsie will 17 | not lose, regardless of how Bessie plays? 18 | 19 | INPUT FORMAT (input arrives from the terminal / stdin): 20 | The first line contains a single integer $T$ ($1 \leq T \leq 10$) representing 21 | the number of test cases. Each test case is described as follows: 22 | First, one line containing three integers $N$, $M$, and $K$, representing 23 | the number of marbles Elsie has, the number of turns, and the number of 24 | potential moves Bessie can make respectively.Then, $M$ lines where 25 | line $i$ contains $K$ distinct space separated integers 26 | $a_{i,1} \; a_{i,2} \ldots a_{i,K}$ ($1 \leq a_{i, j} \leq 10^3$) representing 27 | the possible amounts of marbles that Bessie might play on turn $i$. 28 | It is guaranteed that the sum of $M$ over all test cases is at most 29 | $3 \cdot 10^5$. 30 | 31 | OUTPUT FORMAT (print output to the terminal / stdout): 32 | For each test case, output the lexicographically minimum move sequence for Elsie 33 | to guarantee not losing, or $-1$ if she will lose. The move sequence should be 34 | on a single line and consist of $M$ space-separated tokens each equal to either 35 | "Even" or "Odd". 36 | 37 | Note: "Even" is lexicographically smaller than "Odd". 38 | 39 | SCORING: 40 | Input 3: $M \leq 16$.Inputs 4-6: $M \leq 1000$.Inputs 41 | 7-12: No further constraints. 42 | 43 | 44 | Problem credits: Suhas Nagar 45 | 46 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Problems/index.php?page=viewproblem2&cpid=1421: -------------------------------------------------------------------------------- 1 | 2 | 3 | Farmer John has a permutation $p$ of length $N$ ($2 \leq N \leq 10^5)$, 4 | containing each positive integer from $1$ to $N$ exactly once. However, Farmer 5 | Nhoj has broken into FJ's barn and disassembled $p$. To not be too cruel, FN has 6 | written some hints that will help FJ reconstruct $p$. While there is more than 7 | one element remaining in $p$, FN does the following: 8 | 9 | Let the remaining elements of $p$ be 10 | $p'_1, p'_2, \dots , p'_n$, 11 | If $p'_1 > p'_n$, he writes down $p'_2$ and removes $p'_1$ from the 12 | permutation. Otherwise, he writes down $p'_{n-1}$ and removes $p'_n$ from the 13 | permutation. 14 | At the end, Farmer Nhoj will have written down $N - 1$ integers 15 | $h_1, h_2, \dots, h_{N-1}$, in that order. Given $h$, Farmer John wants to 16 | enlist your help to reconstruct the lexicographically minimum $p$ consistent 17 | with Farmer Nhoj's hints, or determine that Farmer Nhoj must have made a 18 | mistake. Recall that if you are given two permutations $p$ and $p'$, 19 | $p$ is lexicographically smaller than $p'$ if $p_i < p'_i$ at the first 20 | position $i$ where the two differ. 21 | 22 | INPUT FORMAT (input arrives from the terminal / stdin): 23 | Each input consists of $T$ independent test cases ($1\le T\le 10$). Each test 24 | case is described as follows: 25 | 26 | The first line contains $N$. 27 | 28 | The second line contains $N - 1$ integers $h_1, h_2, \dots, h_{N-1}$ 29 | ($1\le h_i\le N$). 30 | 31 | OUTPUT FORMAT (print output to the terminal / stdout): 32 | Output $T$ lines, one for each test case. 33 | 34 | If there is a permutation $p$ of $1\dots N$ consistent with $h$, output the 35 | lexicographically smallest such $p$. If no such $p$ exists, output $-1$. 36 | 37 | SCORING: 38 | Input 2: $N\le 8$Inputs 3-6: $N\le 100$Inputs 7-11: No additional constraints. 39 | 40 | 41 | Problem credits: Chongtian Ma 42 | 43 | -------------------------------------------------------------------------------- /utils/scrapers/prescraped/usaco/Problems/index.php?page=viewproblem2&cpid=1424: -------------------------------------------------------------------------------- 1 | 2 | 3 | **Note: The memory limit for this problem is 512MB, twice the default.** 4 | After years of hosting games and watching Bessie get first place over and over, 5 | Farmer John has realized that this can't be accidental. Instead, he concludes 6 | that Bessie must have winning coded into her DNA so he sets out to find this 7 | "winning" gene. 8 | 9 | He devises a process to identify possible candidates for this "winning" gene. He 10 | takes Bessie's genome, which is a string $S$ of length $N$ where 11 | $1 \leq N \leq 3000$. He picks some pair $(K,L)$ where $1 \leq L \leq K \leq N$ 12 | representing that the "winning" gene candidates will have length $L$ and will be 13 | found within a larger $K$ length substring. To identify the gene, he takes all 14 | $K$ length substrings from $S$ which we will call a $k$-mer. For a given 15 | $k$-mer, he takes all length $L$ substrings, identifies the lexicographically 16 | minimal substring as a winning gene candidate (choosing the leftmost such 17 | substring if there is a tie), and then writes down the $0$-indexed position 18 | $p_i$ where that substring starts in $S$ to a set $P$. 19 | 20 | Since he hasn't picked $K$ and $L$ yet, he wants to know how many candidates 21 | there will be for every pair of $(K,L)$. 22 | 23 | For each $v$ in $1\dots N$, help him determine the number of $(K,L)$ pairs with 24 | $|P|=v$. 25 | 26 | INPUT FORMAT (input arrives from the terminal / stdin): 27 | $N$ representing the length of the string. $S$ representing the given string. 28 | All characters are guaranteed to be uppercase characters where $s_i \in A-Z$ 29 | since bovine genetics are far more advanced than ours. 30 | 31 | OUTPUT FORMAT (print output to the terminal / stdout): 32 | For each $v$ in $1\dots N$, output the number of $(K,L)$ pairs with $|P|=v$, 33 | with each number on a separate line. 34 | 35 | SCORING: 36 | Inputs 2-4: $N \leq 100$ Inputs 5-7: $N \leq 500$ Inputs 8-16: No additional constraints. 37 | 38 | 39 | Problem credits: Suhas Nagar 40 | 41 | -------------------------------------------------------------------------------- /utils/test.py: -------------------------------------------------------------------------------- 1 | from chatbot import Chat 2 | from scrapers import usaco 3 | from prompts import * 4 | 5 | prob = 'https://usaco.org/index.php?page=viewproblem2&cpid=1421' 6 | edi = 'https://usaco.org/current/data/sol_prob3_bronze_open24.html' 7 | 8 | prob_text = usaco.problem(prob) 9 | edi_text = usaco.editorial(prob, edi) 10 | 11 | chatbot = Chat(STEP_GEN(prob_text, edi_text)) 12 | 13 | # query = "I know the last element has to be a 1, because otherwise the minimum element won't be 1 which is impossible. I'm not sure how to proceed from here." 14 | query = "I know that if I am given the first and last element of the original sequence, I can reconstruct it from the hint sequence. I'm not sure where to proceed from here." 15 | 16 | 17 | query += ''' 18 | Only output a list explaining the observations needed from my point in reasoning, even if my line of reasoning not alligned with the official editorial. 19 | The first observation should be my thinking, and then you should build the entire solution from there. Every subsequent observation should flow logically from my thinking. 20 | ''' 21 | print(query) 22 | print(chatbot.chat(query)) --------------------------------------------------------------------------------