{ "info": { "id": "ses_fd018edaeffevjvXwJtcL2L9sb", "slug": "proud-star", "projectID": "global", "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-I0zj28", "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-I0zj28", "title": "Nudge content generation", "agent": "nudge-generation", "model": { "id": "gpt-5.6-sol", "providerID": "openai", "variant": "max" }, "version": "1.18.3", "summary": { "additions": 0, "deletions": 0, "files": 0 }, "cost": 0, "tokens": { "input": 55949, "output": 3667, "reasoning": 10811, "cache": { "read": 59264, "write": 0 } }, "time": { "created": 1787510067793, "updated": 1787510441604 } }, "messages": [ { "info": { "role": "user", "time": { "created": 1787510067811 }, "agent": "nudge-generation", "model": { "providerID": "openai", "modelID": "gpt-5.6-sol", "variant": "max" }, "system": "Follow the user prompt and output schema exactly.", "format": { "type": "json_schema", "schema": { "type": "object", "properties": { "status": { "type": "string", "enum": [ "success", "unsolvable" ], "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource." }, "reason": { "anyOf": [ { "type": "string" }, { "type": "null" } ], "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null." }, "hints": { "anyOf": [ { "type": "array", "description": "Exactly 5 progressive hints, each building on the last.", "items": { "type": "object", "properties": { "order": { "type": "number", "description": "Hint number, 1 through 5." }, "content": { "type": "string", "description": "Markdown hint text." } }, "required": [ "order", "content" ], "additionalProperties": false } }, { "type": "null" } ], "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null." }, "editorial": { "anyOf": [ { "type": "string" }, { "type": "null" } ], "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null." }, "solution": { "anyOf": [ { "type": "string" }, { "type": "null" } ], "description": "If status is 'success', a complete C++ solution. Otherwise null." } }, "required": [ "status", "reason", "hints", "editorial", "solution" ], "additionalProperties": false, "title": "problem_response", "description": "Submit the generated content for the problem, or report if it is unsolvable." }, "retryCount": 2 }, "id": "msg_02fe71263001VNs7XA6YAFu76N", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb" }, "parts": [ { "type": "text", "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nSolve from the supplied statement first. For very hard problems, do a quick source check for the official Codeforces tutorial/editorial and accepted submissions when they would help you derive or verify the solution. If a source is blocked, missing, Cloudflare-challenged, 404, or otherwise unavailable, keep solving from the supplied statement and remember that source status. If you still can't solve after that, it's ok.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2257E: \"Busy Beaver\".\n\nKnown problem metadata:\n- Contest ID: 2257\n- Index: E\n- Name: Busy Beaver\n- Rating: unknown / unrated\n- Tags: brute force, data structures, divide and conquer, dp, greedy, implementation, sortings\n- Problem URL: https://codeforces.com/contest/2257/problem/E\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n\n
E. Busy Beaver
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

The Beaver founded a construction company called \"Busy Beaver\". And now, in order to build up the company's reputation, it needs to construct the tallest building possible.

The company has an initial capital of $$$x$$$ carrots (a very valuable currency for beavers) and $$$n$$$ projects available for construction. A separate site has been allocated for each project, and nothing has been built there yet. A building project is represented by a sequence of contracts for constructing the next floor. To build the $$$j$$$-th floor of the $$$i$$$-th building, it is necessary to spend $$$a_{i, j}$$$ carrots; upon completing it, the company immediately receives $$$b_{i,j}$$$ carrots, which are added to the budget, and the company may use them to build higher floors of the same building or to work on other projects. Since the company is still young, not all contracts are necessarily profitable; in other words, it is possible that $$$a_{i,j} \\gt b_{i,j}$$$.

The Beaver hired you to plan the company's course of action. You choose in which order to build which floors. Note that it is not necessary to complete projects, or even to start them at all. Moreover, between constructing floors of the same building, it is allowed to complete an arbitrary number of contracts unrelated to that building. The main goal is — to build the tallest building possible.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 3 \\cdot 10^4$$$). The description of the test cases follows.

The first line of each input data set contains integers $$$n$$$ and $$$x$$$ — the number of building projects and the initial amount of money ($$$1 \\le n \\le 2 \\cdot 10^5; 0 \\le x \\le 10^{18}$$$).

Next, there are $$$n$$$ descriptions of projects. The first line of the description of project number $$$i$$$ contains the number $$$m_i$$$ — the maximum number of floors available for construction in this building ($$$1 \\le m_i \\le 2 \\cdot 10^5$$$).

The second line of the project description contains $$$m_i$$$ integers $$$a_{i, 1}, a_{i, 2}, \\ldots a_{i, m_i}$$$ — the costs of building the floors. The third line of the project description contains $$$m_i$$$ integers $$$b_{i, 1}, b_{i, 2}, \\ldots b_{i, m_i}$$$ — the profits from building the floors. $$$(0 \\leq a_{i, j}, b_{i, j} \\leq 10^9)$$$.

It is guaranteed that the sum of $$$m_i$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.

Output

For each set of input data, output two numbers — the height, in floors, of the tallest building that can be constructed, and the smallest index of the building for which it is possible to build that number of floors.

Example
Input
2
1 6
4
4 4 2 1
2 4 1 1
2 3
2
4 4
5 5
2
2 20
4 0
Output
4 1
2 1
Note

In the first set, there are enough carrots to sequentially build all $$$4$$$ floors of the only building.

In the second set, you need to first build the first floor of the second building, earning $$$2$$$ carrots from it, after which you can build both floors of the first building.

\n\n\n\nSource lookup status:\n\n- Tutorial (en): https://codeforces.com/blog/entry/156058 loaded (200 OK; title: \"Codeforces Round 1117 (Div. 2) Editorial - Codeforces\"; mentions 2257E)\n\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A deep editorial explaining like literally everything.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be JUST the hint text. No \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nint main() {\n setIO();\n}\n```", "id": "prt_02fe71264001T9UolJ38OUUjzv", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe71263001VNs7XA6YAFu76N" } ] }, { "info": { "parentID": "msg_02fe71263001VNs7XA6YAFu76N", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "max", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-I0zj28", "root": "/" }, "cost": 0, "tokens": { "total": 5474, "input": 2958, "output": 38, "reasoning": 46, "cache": { "write": 0, "read": 2432 } }, "modelID": "gpt-5.6-sol", "providerID": "openai", "time": { "created": 1787510067914, "completed": 1787510084997 }, "finish": "tool-calls", "id": "msg_02fe712ca001h7Qojtfx04h0c3", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb" }, "parts": [ { "type": "step-start", "id": "prt_02fe71cfc001aJ4D7a70470dZx", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe712ca001h7Qojtfx04h0c3" }, { "type": "reasoning", "text": "**Planning immediate tutorial fetching**", "time": { "start": 1787510070528, "end": 1787510071889 }, "metadata": { "openai": { "itemId": "rs_0c4fb22da1b5cbad016a8b3d367b6087d0a9b6bd7d9b4e6517", "reasoningEncryptedContent": "gAAAAABqiz03nX1H0Uil4CfJfJAyx_Bbahi31CgWvlq4djshUO7xgskluo8U8d-jwpMRPaFa1m0_zdtTqfx95KpER7WYHD9kOrYbh7yH89nX_Qlgg9I0RsY8s9jkAuozK0EpUOL78nCzUCQmoKiXJ7-JJCFgDM75QQ48IRan76lYtAZFKAgV471c4IARkC1EjwCh6Uewtnucy92t3SF4UtLRMRCV54SzbEGlsX4SKDZUToEsK7fkc-mp5rfVbXGTIkldmsxiAJvsI-ADy0K730s7AAJFdJ3252rNYW7Ao3mdbeb_DYv-orwOmRiYbdCw9vgdet0aLC3vgBCb_ZmHG1ans5-9oPIVl6sXqQKBxuoJ1Bp8Q4FbHOrQO9cCl0nxr3PjQ8mVbLz1fJwJYzRILgNbcGl_qetJv7Wlu9TZPtmdqUEZc20eO2e8B3cSl-2Ysw4M5-pj81s6kqdUACdYdixmIJmpM6SlpR8qFA4kWJMStW1kYbu8rmZetzt4Y1TqaLO3SK16rpWQT7vs2T4iUjh7m3XxH6QQIxdB7P4FzmSkyINDij3xthe_MQ_K7e4lbP65zyvh7z3eS_GFARZ-vDupfD3IR17r2bDNgoozzlzbTTJHN7z0gblFyyMxJaLVBaKRlPmalwh56JZ4BzEI0Y5O5axXJy2RmEGOe75H1B-M3m_gGuR7prtioGek5yL981qjwu3Y4XAXbqQeHWPGwlLyFQHyvuE3eOwcWEtjOr_1pEnRs3JMxpWPb6IJAZaXWNvx3aH_aSPcI7WJhf70J_cwqQ8ofnKCS2itrVnt97Hf4NUoXiEztpfZYFXywIYtls1uKPlneY_YL5bJ5UBnqH9kGg8WZ4WW9kCNKpPq9vg8jzTmawqwFYZ9Gtmcbev9XxPNz0Od2etjvfKYhjiYzxz1hwYLN26C8Atg3WrHJ4P3UPJgHaExcDi91WQ21krCH8josfcmYx9uouYGWSJ2q51NlCwHo9wxOX7IHDlIP4FZxkvOB0rIBCVrZ8fKl1Yv2Rv_tFKBUhk74Y8n4-qIIWAQ52pykPU-r0vh0i1jORR2CwJ0k4wmGaM_lTj5k7aTTGfY_igUAaS07wb3MSUxNzLd8GrlwTTVofNIAXTOxJLvmQxPmbvhN9tOEbGZM09NGRYrbcjBnEOBAYO1Xi-Evt_KBQ4TnkS9mblnhts1Jd6kwHDWsPzgGZBQOXjEX8QAL8wlIdrZ2QAad86AC7zCOfmg57j_i-3hsGZryel_C7-4kbOOtmYVljXw3Pcnbf0H0sLE4W7wW5s-OWYujCHS8A9JJFrsbbOroOmJj0CRS30EbY5fpFRFJthnpTcC2iiIoBSUhEK3qTqwcpSQWBR3UNKZU0ONasvjmbFi_hUjD5GVTpajmYYsxWnQzsdOUzUAh_Hn_EhhJDps80yhpibyr5KXXGLtt1hCSTEQcfcLBTQYPBIHtf9rb11AyAxUqaRnTxEYRkpHo5QLCZvQI0WGyEengSph2KblHBvI3QjVqpDlxyvUiSeOrqkeR1R-PeIZv8Wol36cHfwhs0RKTm7f287D6iwPeVRQolAWMD2uF6tydjoYa2hPimTEarBBgbRTGkOCWE1qY6lw" } }, "id": "prt_02fe71d00001ILJBgH4w1rcqAr", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe712ca001h7Qojtfx04h0c3" }, { "type": "tool", "tool": "webfetch", "callID": "call_1HucbTDTMIs4kgPM0FZiB2Vr", "state": { "status": "completed", "input": { "url": "https://codeforces.com/blog/entry/156058", "format": "markdown", "timeout": 120 }, "output": "---\nimage: //codeforces.com/codeforces.org/s/99500/images/codeforces-sponsored-by-ton.png\ntitle: Codeforces Round 1117 (Div. 2) Editorial - Codeforces\ndescription: Codeforces. 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Programming Contest](https://codeforces.com/blog/entry/155979) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/99500/images/icons/comment-12x12.png \"New comment(s)\")\n\n| | [Detailed →](https://codeforces.com/recent-actions) |\n| | --------------------------------------------------- |\n\n* [pskobx](https://codeforces.com/profile/pskobx)\n* [Blog](https://codeforces.com/blog/pskobx)\n* [Teams](https://codeforces.com/teams/with/pskobx)\n* [Submissions](https://codeforces.com/submissions/pskobx)\n* [Groups](https://codeforces.com/groups/with/pskobx)\n* [Contests](https://codeforces.com/contests/with/pskobx)\n* [Problemsetting](https://codeforces.com/contests/writer/pskobx)\n\n### [pskobx's blog](https://codeforces.com/blog/pskobx)\n\n[ Codeforces Round 1117 (Div. 2) Editorial ](https://codeforces.com/blog/entry/156058) \n\nBy [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\"), 6 days ago, ![In English](//codeforces.com/codeforces.org/s/99500/images/flags/24/gb.png \"In English\") \n\nThank you for participating in our round! We hope you enjoyed the problems.\n\n[2257A - Creating Abbreviations](https://codeforces.com/contest/2257/problem/A \"Codeforces Round 1117 (Div. 2)\") Idea: [egorka5opka](https://codeforces.com/profile/egorka5opka \"Master egorka5opka\")\n\n**Solution**\n\nNote that adding abbreviations to the set $$$S$$$ does not create any new possibilities for creating new abbreviations, because a word starting with that letter already exists. Therefore, for each of the 26 letters, we can remember that there is a word that begins with it, and then go through all the abbreviations and check each letter.\n\n**Code**\n\n[387467084](https://codeforces.com/contest/2257/submission/387467084 \"Submission 387467084 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257B - Gigantomachy](https://codeforces.com/contest/2257/problem/B \"Codeforces Round 1117 (Div. 2)\") Idea: [egorka5opka](https://codeforces.com/profile/egorka5opka \"Master egorka5opka\")\n\n**Solution**\n\nOn each turn, the height at which the giant stands decreases by exactly $$$1$$$, except in cases where he changes mountains. \n\nThus, the first giant will lose after $$$((a\\_1 - a\\_2) + 1) + ((a\\_2 - a\\_3) + 1) + \\\\dots + ((a\\_{n - 1} - a\\_n) + 1) + a\\_n = a\\_1 + n - 1$$$ moves. \n\nSimilarly, the second giant will lose after $$$b\\_1 + m - 1$$$ moves. All left is only to compare these two numbers.\n\nTime complexity: $$$\\\\mathcal{O}(n)$$$.\n\n**Code**\n\n[387467136](https://codeforces.com/contest/2257/submission/387467136 \"Submission 387467136 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257C - Spying on the Beaver](https://codeforces.com/contest/2257/problem/C \"Codeforces Round 1117 (Div. 2)\") Idea: [egorka5opka](https://codeforces.com/profile/egorka5opka \"Master egorka5opka\")\n\n**Solution**\n\nSuppose we have placed some cameras and remove the corresponding edges from the tree. If all the dams are now in different connected components, we can always uniquely determine where the Beaver went. Conversely, if there is a component containing two dams, we cannot distinguish between them.\n\nSince the original graph is a tree, it is necessary and sufficient to remove $$$m-1$$$ edges.\n\nIf the root contains a dam, we can take the edges leading directly to all the other dams, obtaining a set of size $$$m-1$$$.\n\nIf the root does not contain a dam, we take the same edges but skip one of them. It is important that we cannot skip just any edge: we must choose a dam such that there are no other dams on the path from the root to it. We can take the dam with the smallest depth or vertex number. In this case, we do not even need to run a DFS.\n\nTime complexity: $$$\\\\mathcal{O}(n+m)$$$.\n\n**Code**\n\n[387467226](https://codeforces.com/contest/2257/submission/387467226 \"Submission 387467226 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257D - Bermuda Rectangle](https://codeforces.com/contest/2257/problem/D \"Codeforces Round 1117 (Div. 2)\") Idea: [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\"), [oblememan](https://codeforces.com/profile/oblememan \"Master oblememan\"), [egorka5opka](https://codeforces.com/profile/egorka5opka \"Master egorka5opka\")\n\n**Solution**\n\nBefore processing the queries, let us perform some preprocessing. Find all pairs $$$(a, b)$$$ such that $$$a \\\\cdot b = S$$$. There are $$$\\\\mathcal{O}(\\\\sqrt{S})$$$ such pairs, and they can also be found in $$$\\\\mathcal{O}(\\\\sqrt{S})$$$ time with standard algorithms.\n\nThese pairs are the top-right corners of all possible Bermuda rectangles. The union of these rectangles forms a staircase-like shape; let us call it $$$F$$$. The answer to a query is the area of the intersection of the query rectangle with $$$F$$$.\n\nSort the points by $$$x$$$, and for the point with index $$$i$$$, compute the value $$$P\\_i$$$, equal to the area of $$$F$$$ between $$$0$$$ and this $$$x\\_i$$$.\n\nThis information is enough to answer each query efficiently using binary search. We find the position where the boundary of $$$F$$$ crosses the horizontal side of the query rectangle and calculate the answer using the corresponding prefix areas. The exact formulas and case analysis may vary depending on the implementation.\n\nA couple of remarks that may make implementation easier. If there is a point $$$(x, y)$$$, then there is also $$$(y, x)$$$, and since, knowing one coordinate, it is easy to compute the other. Therefore, it is enough to store just the sorted list of divisors of $$$S$$$ and treat them both as $$$x$$$ and as $$$y$$$. \n\nPreprocessing complexity: $$$\\\\mathcal{O}(\\\\sqrt{S})$$$, time complexity per query: $$$\\\\mathcal{O}(\\\\log S)$$$.\n\n**Code**\n\n[387467301](https://codeforces.com/contest/2257/submission/387467301 \"Submission 387467301 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257E - Busy Beaver](https://codeforces.com/contest/2257/problem/E \"Codeforces Round 1117 (Div. 2)\") Idea: [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\")\n\n**Solution**\n\nThe first step is to earn as many carrots as possible to maximize our capital. Divide the floors of each building into the shortest consecutive segments whose total profit is non-negative, and calculate the entry threshold of each such segment. The entry threshold is the minimum amount of capital required to construct all floors of the segment in order.\n\nFor example, consider two floors with $$$(a\\_1, a\\_2) = (10, 0)$$$ and $$$(b\\_1, b\\_2) = (0, 100)$$$. Constructing only the first floor is unprofitable, but constructing both floors yields a profit of $$$90$$$ carrots. However, an initial capital of at least $$$10$$$ carrots is required.\n\nIn an ordered set, we keep the next available segment for each building, sorted by its entry threshold. While the segment with the minimum threshold is affordable, we construct the entire segment, add its profit to our capital, and insert the next segment of the same building into the set. If the segment with the minimum threshold is not affordable, then none of the available segments can be constructed, meaning that we have maximized our capital.\n\nIn the second step, we independently try to continue constructing each building from its current height, floor by floor, for as long as the current capital allows. Among all buildings, we choose the maximum resulting height and, in case of a tie, the smallest index.\n\nTime complexity: $$$\\\\mathcal{O}\\\\left(\\\\sum m\\_i \\\\log n\\\\right)$$$.\n\n**Code**\n\n[387467400](https://codeforces.com/contest/2257/submission/387467400 \"Submission 387467400 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257F1 - Beaver's Jumping Track (Easy Version)](https://codeforces.com/contest/2257/problem/F1 \"Codeforces Round 1117 (Div. 2)\") Idea: [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\"), [oblememan](https://codeforces.com/profile/oblememan \"Master oblememan\"), [egorka5opka](https://codeforces.com/profile/egorka5opka \"Master egorka5opka\")\n\n**Solution**\n\nSince $$$x$$$ is small, we build a segment tree whose vertices store matrices of size $$$x \\\\times x$$$.\n\nFor an interval of platforms, let $$$c\\[i\\]\\[j\\]$$$ be the minimum penalty if the Beaver starts $$$i$$$ cells after the beginning of the interval and first lands $$$j$$$ cells after its end. Invalid starting positions are ignored.\n\nFor a single platform of length $$$d$$$ and penalty $$$s$$$,\n\n$$$ c\\[i\\]\\[j\\]= \\\\left( \\\\left\\\\lceil\\\\frac{d+j-i}{x}\\\\right\\\\rceil-1 \\\\right)s, $$$\n\nwhere $$$0\\\\le i \\\\lt \\\\min(d,x)$$$ and $$$0\\\\le j \\\\lt x$$$. The last jump leaves the platform and causes no penalty, while all previous jumps stay inside it.\n\nTo merge two consecutive intervals $$$A$$$ and $$$B$$$, enumerate the first landing position $$$k$$$ in $$$B$$$:\n\n$$$ c\\_{A+B}\\[i\\]\\[j\\] = \\\\min\\_k\\\\left(c\\_A\\[i\\]\\[k\\]+c\\_B\\[k\\]\\[j\\]\\\\right). $$$\n\nIf $$$A$$$ contains fewer than $$$x$$$ cells, the Beaver may start directly inside $$$B$$$; in this case, we copy the corresponding values from $$$c\\_B$$$.\n\nFor a query $$$\\[l,r\\]$$$, obtain the matrix for $$$\\[l,r-1\\]$$$ and enumerate the first landing position $$$i$$$ on platform $$$r$$$. The answer is\n\n$$$ \\\\min\\_{0\\\\le i \\\\lt \\\\min(x,d\\_r)} \\\\left( c\\[0\\]\\[i\\] + \\\\left\\\\lceil\\\\frac{d\\_r-i-1}{x}\\\\right\\\\rceil s\\_r \\\\right). $$$\n\nThe case $$$l=r$$$ is calculated directly.\n\nEach update and query takes $$$\\\\mathcal{O}(x^3\\\\log n)$$$ time. Memory complexity is $$$\\\\mathcal{O}(nx^2)$$$. Build is $$$\\\\mathcal{O}(nx^3)$$$.\n\n**Code**\n\n[387467533](https://codeforces.com/contest/2257/submission/387467533 \"Submission 387467533 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2257F2 - Beaver's Jumping Track (Hard Version)](https://codeforces.com/contest/2257/problem/F2 \"Codeforces Round 1117 (Div. 2)\")\n\n**Solution**\n\nSurprisingly, the main problem with the solution from F1 is memory, not time. A segment tree stores $$$\\\\mathcal{O}(n)$$$ matrices, each containing $$$x^2$$$ values. For $$$x=10$$$, this does not fit into the memory limit.\n\nTo reduce memory usage, divide the platforms into blocks of size $$$B$$$ (for example, $$$B=16$$$). For each block, calculate the matrix of the entire block using the same merging formula as in F1\\. Then build a segment tree only over these block matrices.\n\nFor a query, at most two blocks are only partially covered by the required range. Process the platforms in these blocks separately, and use the segment tree to combine all blocks that are fully covered. After an update, recalculate the matrix of the affected block and update it in the segment tree.\n\nFor a block size $$$B$$$, the memory complexity is $$$\\\\mathcal{O}\\\\left(n+\\\\frac{n}{B}x^2\\\\right)$$$.\n\nEach query takes $$$\\\\mathcal{O}\\\\left(Bx^2+x^3\\\\log\\\\frac{n}{B}\\\\right)$$$.\n\nАnd each update takes $$$\\\\mathcal{O}\\\\left(Bx^3+x^3\\\\log\\\\frac{n}{B}\\\\right)$$$.\n\n**Code**\n\n[387467610](https://codeforces.com/contest/2257/submission/387467610 \"Submission 387467610 by pskobx\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nGood [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nOk [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nBad [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\nTerrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n![](//codeforces.com/codeforces.org/s/99500/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1117 (Div. 2)](https://codeforces.com/contest/2257) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* \\-111\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/99500/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/pskobx) [pskobx ](https://codeforces.com/profile/pskobx)\n* ![Publication date](//codeforces.com/codeforces.org/s/99500/images/blog/date_16x16.png \"Publication date\") 6 days ago\n* [![Comments](//codeforces.com/codeforces.org/s/99500/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156058#comments) [209 ](https://codeforces.com/blog/entry/156058#comments)\n\n \n![Comments](//codeforces.com/codeforces.org/s/99500/images/icons/comments-48x48.png \"Comments\") Comments (209) \n\n[Write comment?](#) \n\n| » [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 days ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4963270/avatar/be164ba8d5d4a72e.jpg) ](https://codeforces.com/profile/esc%5Fqwq) [esc\\_qwq](https://codeforces.com/profile/esc%5Fqwq \"Candidate Master esc_qwq\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386304 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +91 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Horrible contest [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [jypo\\_CYL](https://codeforces.com/profile/jypo%5FCYL \"Newbie jypo_CYL\") | 6 days ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ---------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/3846644/avatar/58b83965b1ab953b.jpg) ](https://codeforces.com/profile/jypo%5FCYL) [jypo\\_CYL](https://codeforces.com/profile/jypo%5FCYL \"Newbie jypo_CYL\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386363 \"Link to comment\") [^](#comment-1386304 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) so bad D [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n * | » » » [unorz](https://codeforces.com/profile/unorz \"Newbie unorz\") | 5 days ago, [show (+1)](#) # |\n| ------------------------------------------------------------------ | ---------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4392235/avatar/9e0dd13ad8852cf4.jpg) ](https://codeforces.com/profile/unorz) [unorz](https://codeforces.com/profile/unorz \"Newbie unorz\") | 5 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386715 \"Link to comment\") [^](#comment-1386363 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) interesting, i loved D! Really resembled USACO problems [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n * | » » » » [jypo\\_CYL](https://codeforces.com/profile/jypo%5FCYL \"Newbie jypo_CYL\") | 5 days ago, [show](#) # |\n| -------------------------------------------------------------------------------- | ----------------------- |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3846644/avatar/58b83965b1ab953b.jpg) ](https://codeforces.com/profile/jypo%5FCYL) [jypo\\_CYL](https://codeforces.com/profile/jypo%5FCYL \"Newbie jypo_CYL\") | 5 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386745 \"Link to comment\") [^](#comment-1386715 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) nah it shouldn't be [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n| » [Vctqwq](https://codeforces.com/profile/Vctqwq \"Specialist Vctqwq\") | 6 days ago, [show (+1)](#) # |\n| --------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2479255/avatar/5d52d13c72db55a1.jpg) ](https://codeforces.com/profile/Vctqwq) [Vctqwq](https://codeforces.com/profile/Vctqwq \"Specialist Vctqwq\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386306 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +42 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Super fast editorial! But these statements are difficult to understand :( [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [RizzCoder](https://codeforces.com/profile/RizzCoder \"Newbie RizzCoder\") | 6 days ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5564274/avatar/4212eb081339fb3b.jpg) ](https://codeforces.com/profile/RizzCoder) [RizzCoder](https://codeforces.com/profile/RizzCoder \"Newbie RizzCoder\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386351 \"Link to comment\") [^](#comment-1386306 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) bro editorial is wrong [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [zeyd1234](https://codeforces.com/profile/zeyd1234 \"Specialist zeyd1234\") | 6 days ago, [show](#) # |\n| --------------------------------------------------------------------------- | ----------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4840403/avatar/50b684c8412376e6.jpg) ](https://codeforces.com/profile/zeyd1234) [zeyd1234](https://codeforces.com/profile/zeyd1234 \"Specialist zeyd1234\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386311 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +20 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Bad contest... but at least i gain some rating. [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [\\-rmq](https://codeforces.com/profile/-rmq \"Specialist -rmq\") | 6 days ago, [show (+1)](#) # |\n| ---------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/-rmq) [\\-rmq](https://codeforces.com/profile/-rmq \"Specialist -rmq\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386326 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Can you recheck, if correct code is attached in the editorial. Seems different submission is present here [pskobx](https://codeforces.com/profile/pskobx) [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [RizzCoder](https://codeforces.com/profile/RizzCoder \"Newbie RizzCoder\") | 6 days ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/5564274/avatar/4212eb081339fb3b.jpg) ](https://codeforces.com/profile/RizzCoder) [RizzCoder](https://codeforces.com/profile/RizzCoder \"Newbie RizzCoder\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386333 \"Link to comment\") [^](#comment-1386326 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) yeah I think first has nothing to do with 3 variables [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [zeyd1234](https://codeforces.com/profile/zeyd1234 \"Specialist zeyd1234\") | 6 days ago, [show](#) # |\n| ----------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/4840403/avatar/50b684c8412376e6.jpg) ](https://codeforces.com/profile/zeyd1234) [zeyd1234](https://codeforces.com/profile/zeyd1234 \"Specialist zeyd1234\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386338 \"Link to comment\") [^](#comment-1386326 \"Parent comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) type (user:XXX), if you want to ping someone. But with \\[\\], instead of (). But if you don't want to get your contribution < -100, please, don't ping someone unless necessary. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\") | 6 days ago, [show](#) # |\n| ------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/1250475/avatar/7536d700942a822b.jpg) ](https://codeforces.com/profile/pskobx) [pskobx](https://codeforces.com/profile/pskobx \"Master pskobx\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386346 \"Link to comment\") [^](#comment-1386326 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I'll fix it now, thanks [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [20111019Yu](https://codeforces.com/profile/20111019Yu \"Master 20111019Yu\") | 6 days ago, [show](#) # |\n| ----------------------------------------------------------------------------- | ----------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/3490545/avatar/2b7d9f312416805a.jpg) ](https://codeforces.com/profile/20111019Yu) [20111019Yu](https://codeforces.com/profile/20111019Yu \"Master 20111019Yu\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386329 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) How dare you write the wrong range? [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [bandit07](https://codeforces.com/profile/bandit07 \"Newbie bandit07\") | 6 days ago, [show](#) # |\n| ----------------------------------------------------------------------- | ----------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/5022929/avatar/9f7f959d1770105a.jpg) ](https://codeforces.com/profile/bandit07) [bandit07](https://codeforces.com/profile/bandit07 \"Newbie bandit07\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386332 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Apart from Figuring out TLE on D, everything was great! [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [zhaohongdi114514](https://codeforces.com/profile/zhaohongdi114514 \"Candidate Master zhaohongdi114514\") | 6 days ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------------------------------- | ---------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/zhaohongdi114514) [zhaohongdi114514](https://codeforces.com/profile/zhaohongdi114514 \"Candidate Master zhaohongdi114514\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386334 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +35 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Felt like eating shit. [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [ma\\_niu\\_bi](https://codeforces.com/profile/ma%5Fniu%5Fbi \"Expert ma_niu_bi\") | 6 days ago, [show](#) # |\n| ---------------------------------------------------------------------------------- | ----------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/2710300/avatar/1c5c34a0e21a0f46.jpg) ](https://codeforces.com/profile/ma%5Fniu%5Fbi) [ma\\_niu\\_bi](https://codeforces.com/profile/ma%5Fniu%5Fbi \"Expert ma_niu_bi\") | 6 days ago, [hide](#) [#](https://codeforces.com/blog/entry/156058?#comment-1386672 \"Link to comment\") [^](#comment-1386334 \"Parent comment\") \\| [←](#)Rev. 3 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/99500/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) My brain's full of shit. [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [invoker.\\_](https://codeforces.com/profile/invoker.%5F \"Pupil invoker._\") | 6 days ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ---------------------------- |\n\n\n...359925 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_02fe75573001IaPiBK6yNEs955\nUse Grep to search the full content or Read with offset/limit to view specific sections.", "metadata": { "truncated": true, "outputPath": "/Users/zayd/.local/share/opencode/tool-output/tool_02fe75573001IaPiBK6yNEs955" }, "title": "https://codeforces.com/blog/entry/156058 (text/markdown; charset=utf-8)", "time": { "start": 1787510072081, "end": 1787510084984 } }, "metadata": { "openai": { "itemId": "fc_0c4fb22da1b5cbad016a8b3d37cfa887d08a292b8afaa571bd" } }, "id": "prt_02fe722570011rkuZBM8v934xb", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe712ca001h7Qojtfx04h0c3" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 5474, "input": 2958, "output": 38, "reasoning": 46, "cache": { "write": 0, "read": 2432 } }, "cost": 0, "id": "prt_02fe7557f001puY8mE91CeYGOb", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe712ca001h7Qojtfx04h0c3" } ] }, { "info": { "parentID": "msg_02fe71263001VNs7XA6YAFu76N", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "max", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-I0zj28", "root": "/" }, "cost": 0, "tokens": { "total": 17463, "input": 12881, "output": 76, "reasoning": 26, "cache": { "write": 0, "read": 4480 } }, "modelID": "gpt-5.6-sol", "providerID": "openai", "time": { "created": 1787510084999, "completed": 1787510088945 }, "finish": "tool-calls", "id": "msg_02fe75587001N2F7HsKbqtkRZX", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb" }, "parts": [ { "type": "step-start", "id": "prt_02fe75df9001AmA2gnqgrn2xUX", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe75587001N2F7HsKbqtkRZX" }, { "type": "reasoning", "text": "**Fetching accepted code for inspection**", "time": { "start": 1787510087164, "end": 1787510087483 }, "metadata": { "openai": { "itemId": "rs_0d7409272be49884016a8b3d469f6887d0bd924906394a0581", "reasoningEncryptedContent": "gAAAAABqiz1HjGzIuC45RPd4VjsRtPrNec0_jJHGweTMd5bIpjiSoWocvThHdAKIWqrGNaj2jejxBLU6C9_uCmlYDoIFlgGV7f61Zfyj10-UIHjWX4hSJet8JmdmBwBOEWdBIr8aSMBNCYg_MfI-WyHjeFYChCe-6wXH9O-ywlrvkpUu4DH1pmCj5slbI4HfhNwdNNSBbFESKw3zWnVpygomQcr1j48kb3pDBsyLhtX8hZjxiDq0qWNpy7qboCDmR7sead7e3j4MYp9Gbn1uzHIYWAzlEd-JIC3CStjFHIMx07EqBAjh6uHYZtmD3IvYajVVkQVaFqcaWTkiWBh1WY_B8kQSGkbiLPMwJLb-vDjW6f8p-n2135K9a5EWIA9GN_hAFWeT9wDKsIrrPTt2LcrZHjaAse_mSnCojP4ksxfC-eNBg3JN2MVekZBn3aUWzqaPXq1xIAsTELqbg0VoDK0Vdz8eVQiCqSthd00XrpCBLSFeS7_X6SGIjqTAwOi4TfbfghH3TBrp_fieHatDMoxZOvxPpT4M7B7X8DpxTcvbs5W4Tzgtteq0GgV_aeHI7is2bLtIR56jTU6DOKd-a9Ihvvw0iZecdv8oJUcTT8uSQ8XCifuc8TpUYb6pf31xo4DYb29Z4O0JTRndQ2R6eDU_Gi5gzSCCPbB9wQi-FaUSCHZR_S-kSE-llDziwZSG2FW81SdZPUprpLCyqY5MCg7sAFcydXUepYUZwxIsUYb5YT7spoDlBs9lsi1GjONcUvPlpoj7xa4WB4f_Gr7C4lqTXaUU9Gvu03tglx_ojV0WCs9MpNr0pFyTkBfou0OJHcE_DFYIcZscbTqjSzO1GnEqoLlb8UEJfx7uJmRcgjT5yABYnO5ZTPQggR1nMu4nfO5PLxwcH8a7vhSgNsoqb3mBFblmhj3S1X6XklBglES6SKlBG6HULVNKm2Vw67gqoCo0kuj3-8cLAO_9_0mSVQRz2LKIQBYorEfP6BZZjgJNcnDsGSNXEYiBeT94uwEGL2e9VU8qp7wPkSbK-MBbyrn6qXrqYS2HKG2dCQY3FUb2r6_-mWwk7ozlVTUBhOoYka86FBgK3fDun0ZQoe-Ayrw5q7FvSx0szZ4DaU1usuVESvL9NeIPd8AudaaCyeqjmYQb_UOXP62fd4-vKLzcMzSE87BGa4NaNeaHgBIenyJJeN_CMInOEA2eqqGozZVrLbYhWUzDpUiWY4djQyNf0Df_KwqX4GXxez7BKDRZFf_N2Y1uTYVYprmdMyF_rpEa5H8QZ4epParanG2K9oRkJ8fKwJsSpUyu4hG1NUkEmq43RBkE8gaBR5XyCVk2b7Ihm8tvi1jO6OWW6nHteoF2VVFCokaup8GmR0-5QabDLoGUqeq4Z7e3wWMPilm0dVwHHI58QHA5JgTmi9mVf3AlS0X2PPgvmslDOA==" } }, "id": "prt_02fe75dfc0015mFD8iYguwbjBa", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe75587001N2F7HsKbqtkRZX" }, { "type": "text", "text": "The core greedy is clear; I’m checking the exact segment construction and edge cases against accepted code before writing the proof and implementation.", "time": { "start": 1787510087487, "end": 1787510087913 }, "metadata": { "openai": { "itemId": "msg_0d7409272be49884016a8b3d476db087d0870337bcb1eaf7e5", "phase": "commentary" } }, "id": "prt_02fe75f3f001dVTISd9PuxhNsH", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe75587001N2F7HsKbqtkRZX" }, { "type": "tool", "tool": "webfetch", "callID": "call_3W5RwOObWzOgdAx7LqDwaC5y", "state": { "status": "completed", "input": { "url": "https://codeforces.com/contest/2257/submission/387467400", "format": "markdown", "timeout": 120 }, "output": "Please wait. Your browser is being checked. 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Divide the floors of each building into the shortest consecutive segments whose total profit is non-negative, and calculate the entry threshold of each such segment. The entry threshold is the minimum amount of capital required to construct all floors of the segment in order.\n...\nFor example, consider two floors with $$$(a_1, a_2) = (10, 0)$$$ and $$$(b_1, b_2) = (0, 100)$$$. Constructing only the first floor is unprofitable, but constructing both floors yields a profit of $$$90$$$ carrots. However, an initial capital of at least $$$10$$$ carrots is required.\n...\nIn an ordered set, we keep the next available segment for each building, sorted by its entry threshold. While the segment with the minimum threshold is affordable, we construct the entire segment, add its profit to our capital, and insert the next segment of the same building into the set. If the segment with the minimum threshold is not affordable, then none of the available segments can be constructed, meaning that we have maximized our capital.\n...\nIn the second step, we independently try to continue constructing each building from its current height, floor by floor, for as long as the current capital allows. Among all buildings, we choose the maximum resulting height and, in case of a tie, the smallest index.\n...\nTime complexity: $$$\\mathcal{O}\\left(\\sum m_i \\log n\\right)$$$.\n\n---\n\nTitle: Problemset - Codeforces\nURL: https://codeforces.com/problemset\nPublished: N/A\nAuthor: N/A\nHighlights:\nProblemset - Codeforces\n\n|\n\nProblems\n\nCodeforces(c) Copyright 2010-2026 Mike Mirzayanov\n\nThe only programming contests Web 2.0 platform\n\nServer time: Aug/14/2026 12:51:30 (n1).\n\nDesktop version, switch to mobile version.\n\nPrivacy Policy| Terms and Conditions\n\nSupported by\n\nUser lists\n\n| Name |\n| --- |\n\n---\n\nTitle: Codehunt\nURL: https://codehunt.cc/\nPublished: N/A\nAuthor: N/A\nHighlights:\nCodehunt\n\nCodehunt", "metadata": { "provider": "exa", "truncated": false }, "title": "Exa Web Search: \"2257E\" \"Busy Beaver\" solution C++", "time": { "start": 1787510094127, "end": 1787510095324 } }, "metadata": { "openai": { "itemId": "fc_01c5fb13403486ca016a8b3d4e036c87d0b21ccd79fad69ca3" } }, "id": "prt_02fe7792a001H9Hu50PL90U7T0", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe764f3001IlBdC4ivCq3AfX" }, { "type": "tool", "tool": "websearch", "callID": "call_dqfpkgwsSNMO3P0d2cSIkPz7", "state": { "status": "completed", "input": { "query": "\"387467400\" Codeforces", "numResults": 10, "livecrawl": "preferred", "type": "auto", "contextMaxCharacters": 20000 }, "output": "Title: 火影忍者300集概括图片-西柏林恋曲 Herr Lehmann(2003)-剧大佬\nURL: https://www.zls.co/djbook/880670940/\nPublished: N/A\nAuthor: N/A\nHighlights:\n- 标签： 剧情 / 喜剧 类型：电影导演： 林德·豪尔曼 主演： 评分： 9 剧情： 故事发生在1989年的柏林，莱曼（克里斯蒂安·乌蒙 Christian Ulmen 饰）是一个整天游手好闲无所事事的小混混，沉浸在酒精和毒品之中无法自拔，一事无成。一天，莱曼震惊的得知自己远在他乡的父亲母亲要前来探望他，为了不让父母失望，莱曼找到了好友卡尔（德特勒夫·巴克 Detlev Buck 饰）。 卡尔将自己的餐厅借给了莱曼，莱曼则装成是餐厅的经理，这让莱曼的父母赚足了面子。与此同时，莱曼在餐厅里邂逅了名为卡特琳（卡佳·达诺沃斯基 Katja Danowski 饰）的厨娘，两人之间碰撞出了激情的火花。好景不长，父母以为莱曼混得风生水起，于是将需要接济的亲人都托付给他，这让莱曼叫苦不迭。...\n...\n#### 火影忍者300集概括图片\n...\n- https://www.zls.co/djbook/880670940/主演： 类型： https://www.zls.co/lists/8/地区： 暂无 评分： 2 简介： 信任才需要理由，不信任天经地义。... 评论： 第一集前十五分钟就有0713龙套片段！！！！！！！！！他们六个人演得好好！！！！！ 这部剧get到了邓为，他简直就是天选涂山璟，太适合了，作为纯纯的官配党，磕起来很幸福，第二季快来吧！！！\n...\n8 Magnum", "metadata": { "provider": "exa", "truncated": false }, "title": "Exa Web Search: \"387467400\" Codeforces", "time": { "start": 1787510094135, "end": 1787510095244 } }, "metadata": { "openai": { "itemId": "fc_01c5fb13403486ca016a8b3d4e037887d0956572b81545e6fb" } }, "id": "prt_02fe77935001rO3kk1zUVIdnyL", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe764f3001IlBdC4ivCq3AfX" }, { "type": "tool", "tool": "webfetch", "callID": "call_wHsI0OcBgzSJhggRRwAs4otC", "state": { "status": "completed", "input": { "url": "https://codeforces.com/api/contest.status?contestId=2257&handle=pskobx&from=1&count=100", "format": "text", "timeout": 120 }, "output": 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} }, "id": "prt_02fe85ebb001bkzY2WQOu1O8NU", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe77de4001LPI6Rn6TXaN2mm" }, { "type": "tool", "tool": "webfetch", "callID": "call_j5XbGW6laURMYfwWiWZs3b38", "state": { "status": "completed", "input": { "url": "https://codeforces.com/problemset/submission/2257/387467400", "format": "html", "timeout": 120 }, "output": "\n\n\n\n \n \n\n \n \n \n \n \n\n \n \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n Codeforces\n \n \n \n\n \n \n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n \n \n \n \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n \n \n\n\n \n \n\n \n\n\n\n\n\n\n\n\n
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It is supported only ICPC mode for\n...\ncontests. If you've seen\n...\nproblems, a virtual contest is not for you -\n...\nthese problems in the archive. If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in\n...\narchive. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. \n\n brute force data structures divide and conquer dp greedy implementation sortings No tag edit access\n...\nE. Busy Beaver\n...\nThe Beaver founded a construction company called \"Busy Beaver\". And now, in order to build up the company's reputation, it needs to construct the tallest building possible.\n...\nThe company has an initial capital of $$$x$$$ carrots (a very valuable currency for beavers) and $$$n$$$ projects available for construction. A separate site has been allocated for each project, and nothing has been built there yet. A building project is represented by a sequence of contracts for constructing the next floor. To build the $$$j$$$-th floor of the $$$i$$$-th building, it is necessary to spend $$$a_{i, j}$$$ carrots; upon completing it, the company immediately receives $$$b_{i,j}$$$ carrots, which are added to the budget, and the company may use them to build higher floors of the same building or to work on other projects. Since the company is still young, not all contracts are necessarily profitable; in other words, it is possible that $$$a_{i,j} \\gt b_{i,j}$$$.\n...\nThe Beaver hired you to plan the company's course of action. You choose in which order to build which floors. Note that it is not necessary to complete projects, or even to start them at all. Moreover, between constructing floors of the same building, it is allowed to complete an arbitrary number of contracts unrelated to that building. The main goal is — to build the tallest building possible.\n...\nEach test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 3 \\cdot 10^4$$$). The description of the test cases follows.\n...\nThe first line of each input data set contains integers $$$n$$$ and $$$x$$$ — the number of building projects and the initial amount of money ($$$1 \\le n \\le 2 \\cdot 10^5; 0 \\le x \\le 10^{18}$$$).\n...\nNext, there are $$$n$$$ descriptions of projects. The first line of the description of project number $$$i$$$ contains the number $$$m_i$$$ — the maximum number of floors available for construction in this building ($$$1 \\le m_i \\le 2 \\cdot 10^5$$$).\n...\nThe second line of the project description contains $$$m_i$$$ integers $$$a_{i, 1}, a_{i, 2}, \\ldots a_{i, m_i}$$$ — the costs of building the floors. The third line of the project description contains $$$m_i$$$ integers $$$b_{i, 1}, b_{i, 2}, \\ldots b_{i, m_i}$$$ — the profits from building the floors. $$$(0 \\leq a_{i, j}, b_{i, j} \\leq 10^9)$$$.\n...\nIt is guaranteed that the sum of $$$m_i$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.\n...\nFor each set of input data, output two numbers — the height, in floors, of the tallest building that can be constructed, and the smallest index of the building for which it is possible to build that number of floors.\n...\nIn the first set, there are enough carrots to sequentially build all $$$4$$$ floors of the only building.\n...\nIn the second set, you need to first build the first floor of the second building, earning $$$2$$$ carrots from it, after which you can build both floors of the first building.\n\n---\n\nTitle: Codeforces Round 1117 (Div. 2) Editorial - Codeforces\nURL: https://codeforces.com/blog/entry/156058\nPublished: N/A\nAuthor: N/A\nHighlights:\nBad \n\nTerrible \n\n2257E - Busy Beaver Idea: pskobx\n...\nThe first step is to earn as many carrots as possible to maximize our capital. Divide the floors of each building into the shortest consecutive segments whose total profit is non-negative, and calculate the entry threshold of each such segment. The entry threshold is the minimum amount of capital required to construct all floors of the segment in order.\n...\nFor example, consider two floors with $$$(a_1, a_2) = (10, 0)$$$ and $$$(b_1, b_2) = (0, 100)$$$. Constructing only the first floor is unprofitable, but constructing both floors yields a profit of $$$90$$$ carrots. However, an initial capital of at least $$$10$$$ carrots is required.\n...\nIn an ordered set, we keep the next available segment for each building, sorted by its entry threshold. While the segment with the minimum threshold is affordable, we construct the entire segment, add its profit to our capital, and insert the next segment of the same building into the set. If the segment with the minimum threshold is not affordable, then none of the available segments can be constructed, meaning that we have maximized our capital.\n...\nIn the second step, we independently try to continue constructing each building from its current height, floor by floor, for as long as the current capital allows. Among all buildings, we choose the maximum resulting height and, in case of a tie, the smallest index.\n...\nTime complexity: $$$\\mathcal{O}\\left(\\sum m_i \\log n\\right)$$$.\n...\nRate the problem\n\n---\n\nTitle: Codeforces #1117 : Div2. E. Busy Beaver - kmjp's blog\nURL: https://kmjp.hatenablog.jp/entry/2026/08/20/0930\nPublished: 2026-08-20T00:00:00.000Z\nAuthor: kmjp\nHighlights:\nCodeforces #1117 : Div2. E. Busy Beaver - kmjp's blog\n\n 2026- 08- 20 \n\n# Codeforces #1117 : Div2. E. Busy Beaver \n\nこれは途中ミスこそあったものの、解法自体はすんなり出せたな。 https://codeforces.com/contest/2257/problem/E\n...\nプレイヤーは初期状態でXのお金を持つ。\n...\nここで、N個の建物を建てるプランがあり、各建物には異なる階数が設定されていて、階ごとに建築コストと立てたあとの利益が設定されている。\n...\nある階を建築するには、当然下の階を先に建築する必要がある。 また、コスト以上のお金を所有している必要があり、建築完了すると即座に利益分のお金が手に入る。 建築の順番は任意で、途中他の建物に移ってもよい。\n...\n最も高い建物を建てるられるのは、どの建物で何回か。\n...\n### 解法\n...\n各建物を最大まで高くする場合を総当たりする。 それに先立ち、まずお金を増やすことはどの建物を建てる場合も有効なのでそれを行っておこう。 建物ごとにコストと利益の累積和を考え、利益の方が多くなるタイミングがあれば、コストの低い順に適宜採用して建築していく。\n\nあとは、それ以上お金が増えることはないので、各建物を総当たりして全額betし、どこまで高くできるか求める。\n...\n```cpp\nint T;\nint N;\nll X;\nint M[202020];\nvector A[202020],B[202020];\nvector> V[202020];\nint num[202020];\n...\nvoid solve() {\n\tint i,j,k,l,r,x,y; string s;\n\t\n\tcin>>T;\n\twhile(T--) {\n\t\tcin>>N>>X;\n\t\tpriority_queue> Q;\n\t\tFOR(i,N) {\n\t\t\tcin>>M[i];\n\t\t\tA[i].resize(M[i]);\n\t\t\tFORR(a,A[i]) cin>>a;\n\t\t\tB[i].resize(M[i]);\n\t\t\tFORR(a,B[i]) cin>>a;\n\t\t\t\n\t\t\tV[i].clear();\n\t\t\tll sum=0,mi=0;\n\t\t\tnum[i]=0;\n\t\t\tFOR(j,M[i]) {\n\t\t\t\tsum-=A[i][j];\n\t\t\t\tmi=min(mi,sum);\n\t\t\t\tsum+=B[i][j];\n\t\t\t\tif(sum>0) {\n\t\t\t\t\tV[i].push_back({mi,sum,i,j+1});\n\t\t\t\t\tsum=mi=0;\n\t\t\t\t}\n\t\t\t}\n\t\t\tif(V[i].size()) {\n\t\t\t\tQ.push(V[i][0]);\n\t\t\t}\n\t\t}\n\t\twhile(Q.size()) {\n\t\t\tauto p=Q.top();\n\t\t\tQ.pop();\n\t\t\tif(X+p[0]<0) break;\n\t\t\tX+=p[1];\n\t\t\ti=p[2];\n\t\t\tnum[i]++;\n\t\t\tif(num[i]ma) {\n\t\t\t\tma=floor;\n\t\t\t\tret=i+1;\n\t\t\t}\n\t\t}\n\t\tcout< 0), you use a min-heap to repeatedly check if your current total money M meets the cheapest available block's requirement (M≥ R), and if so, you build it and update your wealth to M n e w= M+ P. Finally, having reached your mathematical maximum wealth from these profitable groups, you simulate building the remaining individual floors one by one by checking if M≥ A i and updating your money to M n e w= M− A i+ B i while simply tallying the total floors built for each building to find the absolute maximum height.\n\n---\n\nTitle: If you are trying to understand the optimal strategy for E - Codeforces\nURL: https://codeforces.com/topic/156717/en1\nPublished: N/A\nAuthor: N/A\nHighlights:\nIf you are trying to understand the optimal strategy for E - Codeforces\n...\nIf you are trying to understand the optimal strategy for E\n\n Revision en1, by IBACKFORREVENGE, 2026-08-17 21:50:24 \n\nIf you are trying to understand the optimal strategy for E.\n...\nBusy Beaver, you can think of it like an RPG city-building game where your goal is to use greedy block merging and a min-heap to grind for cash and build the tallest skyscraper possible. You start with a small amount of carrots and need to complete profitable construction \"quests\" across various sites before dumping all your accumulated wealth into a single building. Since some individual floors make a profit while others lose money, you first merge any unprofitable floor with the floors directly above it until the combined block yields a strictly positive net profit, creating a \"quest\" with a specific peak money requirement to survive construction and a total net payout at the end. Once you have these strictly profitable quests, you throw the first available one from every building into a min-heap sorted by their entry requirement and greedily complete the cheapest ones you can afford. Because every quest is profitable, your bank account snowballs, allowing you to unlock and add the next quests for those buildings into the heap until you are either too poor to afford the remaining ones or the heap is entirely empty. Having reached the absolute maximum wealth possible in the game, you then take your massive pile of cash and simulate buying the remaining unbuilt floors of each partially-constructed building one by one until you go completely broke, simply keeping track of which building reaches the highest floor. \n\nThis entire process efficiently groups the floors and processes the optimal path in $$$O(\\sum m_i \\log N)$$$ time, which easily passes the two-second time limit!\n...\nTo mathematically solve the \"Busy Beaver\" problem, you must model the construction process using four core formulas applied in a single continuous workflow. First, every individual floor $$$i$$$ has an upfront cost $$$A_i$$$ and a completion reward $$$B_i$$$, meaning its net profit is calculated as $$$P = B_i - A_i$$$ and its base money requirement to start construction is $$$R = A_i$$$. Because some floors lose money ($$$P \\le 0$$$), you must mathematically merge them with the floors directly above them into a combined block; when joining a lower block (with requirement $$$R_1$$$ and profit $$$P_1$$$) to an upper block (with requirement $$$R_2$$$ and profit $$$P_2$$$), the combined net profit is simply the sum $$$P_{new} = P_1 + P_2$$$, while the new combined minimum requirement becomes $$$R_{new} = \\max(R_1, R_2 - P_1)$$$ because you need enough starting capital to afford the first block and also enough to cover the second block's cost after accounting for the first block's net cash change. Once all floors are merged into strictly positive profit blocks ($$$P \\gt 0$$$), you use a min-heap to repeatedly check if your current total money $$$M$$$ meets the cheapest available block's requirement ($$$M \\ge R$$$), and if so, you build it and update your wealth to $$$M_{new} = M + P$$$. Finally, having reached your mathematical maximum wealth from these profitable groups, you simulate building the remaining individual floors one by one by checking if $$$M \\ge A_i$$$ and updating your money to $$$M_{new} = M - A_i + B_i$$$ while simply tallying the total floors built for each building to find the absolute maximum height.\n\n---\n\nTitle: If you are trying to understand the optimal strategy for E - Codeforces\nURL: https://codeforces.com/topic/156717/diff/en1/en2\nPublished: N/A\nAuthor: N/A\nHighlights:\nIf you are trying to understand the optimal strategy for E - Codeforces\n...\nIf you are trying to understand the optimal strategy for E \n\n Difference between en1 and en2, changed 50 character(s)\n...\n[**If you are trying to understand the optimal strategy for E.**](https://codeforces.com/contest/2257/problem/E)↵ ↵ Busy Beaver, you can think of it like an RPG city-building game where your goal is to use greedy block merging and a min-heap to grind for cash and build the tallest skyscraper possible. You start with a small amount of carrots and need to complete profitable construction \"quests\" across various sites before dumping all your accumulated wealth into a single building. Since some individual floors make a profit while others lose money, you first merge any unprofitable floor with the floors directly above it until the combined block yields a strictly positive net profit, creating a \"quest\" with a specific peak money requirement to survive construction and a total net payout at the end. Once you have these strictly profitable quests, you throw the first available one from every building into a min-heap sorted by their entry requirement and greedily complete the cheapest ones you can afford. Because every quest is profitable, your bank account snowballs, allowing you to unlock and add the next quests for those buildings into the heap until you are either too poor to afford the remaining ones or the heap is entirely empty. Having reached the absolute maximum wealth possible in the game, you then take your massive pile of cash and simulate buying the remaining unbuilt floors of each partially-constructed building one by one until you go completely broke, simply keeping track of which building reaches the highest floor. ↵ ↵ ↵ This entire process efficiently groups the floors and processes the optimal path in $O(\\sum m_i \\log N)$ time, which easily passes the two-second time limit!↵ ↵ ↵ To mathematically solve the \"Busy Beaver\" problem, you must model the construction process using four core formulas applied in a single continuous workflow. First, every individual floor $i$ has an upfront cost $A_i$ and a completion reward $B_i$, meaning its net profit is calculated as $P = B_i - A_i$ and its base money requirement to start construction is $R = A_i$. Because some floors lose money ($P \\le 0$), you must mathematically merge them with the floors directly above them into a combined block; when joining a lower block (with requirement $R_1$ and profit $P_1$) to an upper block (with requirement $R_2$ and profit $P_2$), the combined net profit is simply the sum $P_{new} = P_1 + P_2$, while the new combined minimum requirement becomes $R_{new} = \\max(R_1, R_2 - P_1)$ because you need enough starting capital to afford the first block and also enough to cover the second block's cost after accounting for the first block's net cash change. Once all floors are merged into strictly positive profit blocks ($P > 0$), you use a min-heap to repeatedly check if your current total money $M$ meets the cheapest available block's requirement ($M \\ge R$), and if so, you build it and update your wealth to $M_{new} = M + P$. Finally, having reached your mathematical maximum wealth from these profitable groups, you simulate building the remaining individual floors one by one by checking if $M \\ge A_i$ and updating your money to $M_{new} = M - A_i + B_i$ while simply tallying the total floors built for each building to find the absolute maximum height.↵ ↵ ↵ Love cp \n\nby — [user:IBACKFORREVENGE,2026-08-17] ↵ ↵ \n\n#### History\n\n Revisions \n\n \n\n| Rev. | Lang. | By | When | Δ | Comment |\n| --- | --- | --- | --- | --- | --- |\n| en2 | | IBACKFORREVENGE | 2026-08-17 21:52:39 | 50 | Tiny change: '\\nLove cp ' -> '\\nLove cp by - [user:IBACKFORREVENGE] \\n\\n' |\n| en1 | | IBACKFORREVENGE | 2026-08-17 21:50:24 | 3362 | Initial revision (published) |\n\n---\n\nTitle: Solutions to Codeforces problems - GitHub\nURL: https://github.com/deepakkasera/CodeForces\nPublished: N/A\nAuthor: N/A\nHighlights:\n# deepakkasera/CodeForces\n...\nSolutions to Codeforces problems \n\n- Stars: 0\n- Forks: 0\n- Watchers: 0\n- Open issues: 0\n- Homepage: http://codeforces.com/\n- Default branch: master\n- Created: 2016-10-25T18:45:38Z\n\n---\n\nTitle: Codeforces-Solution - GitHub Pages\nURL: https://fahimfba.github.io/Codeforces-Solution/\nPublished: N/A\nAuthor: N/A\nHighlights:\nCodeforces-Solution | My solutions from Codeforces. 💻 Skip to the content.\n...\n# Codeforces Solution\n\n---\n\nTitle: Search code, repositories, users, issues, pull requests...\nURL: https://github.com/Abritty/Codeforces\nPublished: 2020-07-29T00:00:00.000Z\nAuthor: Abritty\nHighlights:\n|\n|\n###\n...\n[356. Bear & Criminals(B).cpp](https://github.com/Abritty/Codeforces/blob/master/356.%20Bear%20&%20Criminals(B).cpp)\n|\n###\n...\nhttps://github\n...\nCodeforces/blob/\n\n---\n\nTitle: Search code, repositories, users, issues, pull requests...\nURL: https://github.com/sensssey/codeforces\nPublished: 2023-04-07T00:00:00.000Z\nAuthor: sensssey\nHighlights:\n[9 Commits](https://github.com/sensssey/codeforces/commits/algorithms/)\n[](https://github.com/sensssey/codeforces/commits/algorithms/)\n|\n|\n[LCA.cpp](https://github.com/sensssey/codeforces/blob/algorithms/LCA.cpp)\n|\n[LCA.cpp](https://github.com/sensssey/codeforces/blob/algorithms/LCA.cpp)\n|\n \n|\n \n|\n|\n[algorithm\\_Mo.cpp](https://github.com/sensssey/codeforces/blob/algorithms/algorithm_Mo.cpp)\n|\n[algorithm\\_Mo.cpp](https://github.com/sensssey/codeforces/blob/algorithms/algorithm_Mo.cpp)\n|\n \n|\n \n|\n|\n[bfs.cpp](https://github.com/sensssey/codeforces/blob/algorithms/bfs.cpp)\n|\n[bfs.cpp](https://github.com/sensssey/codeforces/blob/algorithms/bfs.cpp)\n|\n \n|\n \n|\n|\n[dijkstra.cpp](https://github.com/sensssey/codeforces/blob/algorithms/dijkstra.cpp)\n|\n[dijkstra.cpp](https://github.com/sensssey/codeforces/blob/algorithms/dijkstra.cpp)\n|\n \n|\n \n|\n|\n[prefix\\_function.cpp](https://github.com/sensssey/codeforces/blob/algorithms/prefix_function.cpp)\n|\n[prefix\\_function.cpp](https://github.com/sensssey/codeforces/blob/algorithms/prefix_function.cpp)\n|\n \n|\n \n|\n|\n[segment\\_tree.cpp](https://github.com/sensssey/codeforces/blob/algorithms/segment_tree.cpp)\n|\n[segment\\_tree.cpp](https://github.com/sensssey/codeforces/blob/algorithms/segment_tree.cpp)\n|\n \n|\n \n|\n|\n[z-function.cpp](https://github.com/sensssey/codeforces/blob/algorithms/z-function.cpp)\n|\n[z-function.cpp](https://github.com/sensssey/codeforces/blob/algorithms/z-function.cpp)\n|\n \n|\n \n|\n|\nView all files\n|\n## About\nNo description, website, or topics provided.\n[", "metadata": { "provider": "exa", "truncated": false }, "title": "Exa Web Search: github Codeforces 2257 E Busy Beaver 2257E.cpp", "time": { "start": 1787510155165, "end": 1787510156235 } }, "metadata": { "openai": { "itemId": "fc_0492323e0977dd17016a8b3d8afe2887d0bb8d884820e1f970" } }, "id": "prt_02fe8679a001n1hj3R2JznFgBg", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe77de4001LPI6Rn6TXaN2mm" }, { "type": "tool", "tool": "websearch", "callID": "call_xgvLN62jhAULzIo4IDKwHe72", "state": { "status": "completed", "input": { "query": "\"Busy Beaver\" \"entry threshold\" Codeforces 2257E", "numResults": 10, "livecrawl": "preferred", "type": "deep", "contextMaxCharacters": 30000 }, "output": "Title: Codeforces Round 1117 (Div. 2) Editorial - Codeforces\nURL: https://codeforces.com/blog/entry/156058\nPublished: N/A\nAuthor: N/A\nHighlights:\nTerrible\n...\n2257E - Busy Beaver Idea: pskobx\n...\nThe first step is to earn as many carrots as possible to maximize our capital. Divide the floors of each building into the shortest consecutive segments whose total profit is non-negative, and calculate the entry threshold of each such segment. The entry threshold is the minimum amount of capital required to construct all floors of the segment in order.\n...\nFor example, consider two floors with $$$(a_1, a_2) = (10, 0)$$$ and $$$(b_1, b_2) = (0, 100)$$$. Constructing only the first floor is unprofitable, but constructing both floors yields a profit of $$$90$$$ carrots. However, an initial capital of at least $$$10$$$ carrots is required.\n...\nIn an ordered set, we keep the next available segment for each building, sorted by its entry threshold. While the segment with the minimum threshold is affordable, we construct the entire segment, add its profit to our capital, and insert the next segment of the same building into the set. If the segment with the minimum threshold is not affordable, then none of the available segments can be constructed, meaning that we have maximized our capital.\n...\nIn the second step, we independently try to continue constructing each building from its current height, floor by floor, for as long as the current capital allows. Among all buildings, we choose the maximum resulting height and, in case of a tie, the smallest index.\n...\nTime complexity: $$$\\mathcal{O}\\left(\\sum m_i \\log n\\right)$$$.\n\n---\n\nTitle: When Will This End?\nURL: https://cacm.acm.org/news/when-will-this-end/\nPublished: 2025-03-11T00:00:00.000Z\nAuthor: Chris Edwards\nHighlights:\nThe Busy Beaver Challenge group, founded by Tristan Stérin, a computer science researcher at the University of Maynooth in Ireland, in early 2022 coordinated their results through a combination of the Discord chat service, a wiki, and Github’s code repository. With the help of these collaboration tools, the group built on existing work by people working mostly independently and supplying results to a central repository. In doing so, they attacked one of the oldest problems in computer science: how to work out when, and indeed whether, any program will terminate.\n...\nResearchers quickly found as the number of program states increases, the length of the printed tape increases at an extraordinarily rapid pace. A binary machine that can switch between two states, which implies a four-entry truth table, prints no more than six bits unless its program winds up in an infinite loop. If you move to a four-state machine, denoted as BB(4,2), the longest-program will not stop before printing a hundred bits. With BB(5,2), the addition of a fifth state pushed this way further. The Busy Beaver Challenge group proved in 2024 it stands at 41,176,870.\n...\nThanks to the automated tools, the raw number of holdouts yet to be analyzed in depth were more than halved from around 12,000 in less than six months after the team moved on from the completed BB(5,2). However, even when they are not cryptids, holdouts in this class of machine present additional levels of difficulty. The current reigning champion for BB(6,2), discovered by Kropitz, is a machine that could print a tape with more digits than there are particles in the known universe before halting. But it proved amenable to a hand-generated proof.\n...\nThe open question for beaver hunters is whether BB(6,2) will yield a definite answer. Michel expects the situation to lead to the situation where there will be a strong candidate for champion, but only on the condition that known cryptids the group and others find are assumed to never halt. A more definite result, however, would mean an incredible breakthrough in mathematics because it would shed a new light on the Collatz conjecture at the very least.\n\n---\n\nTitle: mfornet/busybeaver\nURL: https://github.com/mfornet/busybeaver\nPublished: N/A\nAuthor: N/A\nHighlights:\nThis repository contains a Lean 4 formalization of the five-state, two-symbol\nBusy Beaver result:\n...\n> **BB(5) = 47,176,870**\n...\nHere `BB(5)` is the maximum number of transitions made before halting by a\nhalting five-state, two-symbol Turing machine started on a blank tape. The Lean\nlibrary internally counts steps to the pre-halt configuration, so its\ncorresponding theorem is `Busybeaver 4 1 = 47,176,869`; the theorem\n`BBTheorems.bb5_literature` states the standard value above.\n...\nWhen no configuration file is passed, the executable uses a size-aware default:\n`BB(3,2)` includes the history-augmented NGram CPS pass needed to close the\nknown holdouts, `BB(4,2)` uses the complete Coq-style pipeline, `BB(5,2)` uses\nthe Coq NGram pipeline plus the generated hardcoded table, and other larger runs\nuse a lighter development default.\n...\nFor `BB(\n...\n,2)`, the default follows\n...\n(4)` pipeline\n...\nthe Coq proof\n...\nloop detection with\n...\n`107`, the standard NGram CPS passes, the fixed-history\n...\nand LRU NGram CPS passes\n...\n`{ \"len\": 4, \"threshold\": 3,\n...\n\"maxT\": 320, \"bound\": 10000 }\n...\nFor `BB(5,2)`, the default uses bounded exploration passes for partial-machine\nexpansion; the Loop1, NGram CPS, LRU, and RepWL passes from the Coq BB5\npipeline for which Lean has executable equivalents; and `\"bb5TableExecutable\"`.\n...\nThe generated table contains all 8,228 hardcoded Coq rows, including custom\nNGram, RepWL, halt, Loop1, FAR, WFAR, and sporadic entries.\n...\nThe concrete values `BB(2,2)` … `BB(5,2)` are stated as Lean theorems in the\n[BBTheorems](./BBTheorems/) library (for example,\n`BBTheorems.bb4 : Busybeaver 3 1 = 106`,\nwith a `_literature` companion in the convention that counts the halting\ntransition, `Busybeaver 3 1 + 1 = 107`). Each theorem instantiates\n`Busybeaver.BBCompute.correct_complete` with the CLI's decider pipeline and\ndischarges the \"no undecided machines\" hypothesis by `native_decide`.\n...\nof the BB5 formalisation builds\n...\nq-BB\n...\n](https://github.com\n...\nccz181078/\n...\nproof that `BB(\n...\n47,176,870\n...\nfrom which the deciders\n...\nthe hardcoded\nmachine table,\n\n---\n\nTitle: Shtetl-Optimized\nURL: https://scottaaronson.blog/?p=7489\nPublished: 2023-08-21T00:00:00.000Z\nAuthor: N/A\nHighlights:\n2. Duane Rich created a Part II of his YouTube video series on the Busy Beaver function. It features some of the core ideas from my Busy Beaver survey, clearly narrated and beautifully animated. If reading my survey is too much for you, now you can just watch the movie!\n...\n@Scott 26: Yes, coding is the one major exception (ironically, one may add), though LLMs are still mostly assisting humans, and are still poor at solving totally novel tasks beyond their training data (the CodeForce example).\n...\n@Hans Holdander #34: You mention the Codeforces problems earlier in this thread as examples where the model is poor at solving “totally novel tasks”. It’s hard to quantify what one means by “totally novel tasks”. e.g. One could take a very bright high school student who is “good at math”, and then give them the IMO exam (or the Putnam exam), and maybe they totally fail, maybe even getting not a single problem correct. Would we then conclude that the student must not be able to solve totally novel problems at all? I mean here we gave them tests that didn’t even require any advanced math!\n...\nMy initial reaction\n...\nnot surprising.\n...\nfirst year’\n...\nwere probably in\n...\nthese are competition problems, it\n...\non the next year’s ones\n...\nhadn’t seen.” (At the\n...\nsome of these\n...\ncompetition problems require knowing some mathematical tricks, so can be like IMO problems. e.g. that if p is a prime number and x is an integer in [1,p-1], then the mapping f(y) : {0,…,p-1} –> {0,…,p-1} sending y to xy mod p is a permutation of {0,…,p-1}; and some problem might even require leveraging the fact that for p large enough not every permutation of {0,…,p-1} is of this form.)\n\n---\n\nTitle: Busy beaver - Ciro Santilli (@cirosantilli) - OurBigBook.com\nURL: https://ourbigbook.com/cirosantilli/busy-beaver\nPublished: 2023-10-27T00:00:00.000Z\nAuthor: Ciro Santilli ( @cirosantilli, 40  )\nHighlights:\n23, it had basically been decided\n...\nthe The Busy Beaver Challenge\n...\nmentioned at: discuss.bbchallenge.org/t/the-30-to\n...\n34-ctl-\n...\nouts-from-\n...\n-5/141, pending only\n...\nverification. It is\n...\nfinally settle.\n...\nEntry on The Busy Beaver Challenge: bbchallenge.org/1RB1LC_1RC1RB_1RD0LE_1LA1LD_1RZ0LA\n...\nBut maybe, just maybe, reduction to Turing machine form could be useful. E. g. The Busy Beaver Challenge and other attempts to solve BB(5) have come up with large number of automated (usually parametrized up to a certain threshold) Turing machine decider programs that automatically determine if certain (often large numbers of) Turing machines run forever.\n...\nThe \" Busy Beaver\" is a concept in computability theory and theoretical computer science that relates to Turing machines, which are abstract mathematical models of computation. The Busy Beaver function, often denoted as\\( BB(n) \\), is defined for a Turing machine with \\(n\\) states that halts on all possible inputs. The function gives the maximum number of non-blank symbols that such a Turing machine can output before halting.\n\n---\n\nTitle: http://www.rpi.edu/∼\nURL: https://www.newdualism.org/papers/S.Bringsjord/laihyper11.pdf\nPublished: N/A\nAuthor: N/A\nHighlights:\n” (or Σ\n...\ncan hypercompute\n...\nis intended to\n...\n5.1 The Problem Defined \nThe “busy beaver” or Σ function is a mapping from N to N such that: Σ(n) is the largest number \nof contiguous 1’s that an n-state Turing machine with alphabet { 0, 1} can write on its initially \nblank tape, just before halting with its read/write head on the leftmost 1, where a sequence \nm times \nz }| { 11 . . . 11 \nis regarded simply as m. 9 Rado (1963) proved this function to be Turing-uncompuable long ago; a \nnice contemporary version of the proof (which is by the way not based on diagonalization) is given \nin (Boolos & Jeffrey 1989). Nonetheless, the busy beaver problem is the challenge of determining \nΣ(n) for ever larger values of n. \nOf course, there are varying definitions of Turing machines. The formalism we prefer is the \nquadruple transition, implicit halt state one. In this scheme, each transition consists of four things: \nthe state the machine is in, the symbol it’s scanning, the action it is to perform (move right or \nleft, or write a symbol), and the new state it is to enter. To make it clear that this is the specific \nscheme we’re talking about, we write Σ B, where the B is in honor of the excellent presentation in \n(Boolos & Jeffrey 1989). \n5.2 RAIR Lab Approach Hitherto; Records \nWith\n...\n1983, Machlin & Stout 1990, Marxen & Buntrock 1990), we have defined the following algorithm \nas a solution to Σ B(n): \n1. Using a tree normalized approach, 10 enumerate a set S of n-state Turing machines that behaviorally \nrepresents the entire set. \n2. For each member t of S: \n(a) Classify t as either a non-halter, or a halter \n(b) If t is a halter, run t to completion on an infinite bi-directional tape of all 0’s. If the resulting \ntape satisfies the conditions specified in section 5.1, add t to our candidate set C. \n3. Return the productivity of the most productive machine in C (i.e. the machine that produces the \nmost contiguous 1’s on the tape) \nClearly, the task in item 2a is menacing. While we can’t hope to define a computational \nalgorithm to compute this task (as this would be a solution to the halting problem below the \nTuring limit), we can still strive to prove that a particular machine does not halt on an individual \nbasis by demonstrating that its behavior follows an infinite, repeatable pattern. \nThis technique, pioneered in (Lin & Rado 1964), and expanded upon in (Brady 1983, Machlin\n...\n1990), had previously produced a number of proven non-halter detection strategies and \nbehaviors. Each of these behaviors encompass some subset of non-halting Turing machines within\n...\nscope of all non-halting Turing machines relevant to Σ B(n). The classifications range from \nTuring machines that can be proven non-halters by working backwards from their required halting \nconditions, to simple recurrence patterns, to even machines that mimic an abstraction of binary \ncounters.\n...\na case in point, let us examine one of\n...\npatterns: so-called “Christmas Trees,” discussed\n...\n0). For purposes\n...\nthis discussion, consider\n...\nfollowing \nnotation used\n...\n0∗[U][X][X][X\n...\nt exhibits one of\n...\nhalting behaviors\n...\nto a predetermined step limit on an infinite bi\n...\ndirectional tape of all 0’s. (c) If t halts before or at this step limit, add\n...\nset C if the\n...\nhas not halted after\n...\nlimit has been reached, add it to\n...\n6 The New G¨odelian Argument First, note that, for generality, we drop the subscript B from Σ B. Then here we go: Proof. The argument is at its core a reductio. The following constitutes the argument’s premises:\n...\n• There exists a human person who has determined the productivity of the initial segment of Turing machines (such people are in the RAIR Lab): (1) ∃ p(D(p, Σ(1)) ∧ . . . ∧ D(p, Σ(6)))\n...\n• There is a natural number at and beyond which Turing machines of size less than or equal to k fail (with respect to determining productivity): (2) ∃ n∀ m(#(m) ≤ k → ¬ D(m, Σ(n)) ∧ ¬ D(m, Σ(n + 1)) ∧ ¬ D(m, Σ(n + 2)). . .)\n...\n• If a person can determine the productivity for n, he/she can determine it for n + 1: (3) ∀ n∀ p(D(p, Σ(n)) → D(p, Σ(n + 1))) \nNow, to generate absurdity, suppose that computationalism holds, i.e., reiterating from above: \nC 0 ∀ p∃ m(p = m ∧ #(m) ≤ k) \nNext, suppose that p ? is an arbitrary person who determines the initial segment of the busy \nbeaver problem, i.e., \n\n(ρ) (D(p ?, Σ(1)) ∧ . . . ∧ D(p ?, Σ(6))\n...\nBy universal elimination on (C 0) we have: \n(4) ∃ m(p ? = m ∧ #(m) ≤ k) \nNext, we make the supposition, with m ? as arbitrary, that \n(5) (p ? = m ? \n∧ #(m ?) ≤ k) \nand likewise the supposition, with n ? arbitrary, that \n\n(6) ∀ m(#(m) ≤ k → ¬ D(m, Σ(n ?)) ∧ ¬ D(m, Σ(n ? + 1)) ∧ ¬ D(m, Σ(n ? + 2). . .)\n...\nfrom which it follows by universal elimination that (7) (#(m ?) ≤ k → ¬ D(m ?, Σ(n ?)) ∧ ¬ D(m ?, Σ(n ? + 1)) ∧ ¬ D(m ?, Σ(n ? + 2). . .)\n...\nFrom (5) and (7) we can infer\n...\n¬ D(m ?, Σ(n ?)) ∧ ¬ D(m ?, Σ(n ? + 1)) ∧ ¬ D(m ?, Σ(n ? + 2). . .) but by identity elimination and induction using (3), (5), and (ρ), we can deduce ∀ nD(m ?, Σ(n)) — contradiction. It follows that since (i) (as we’ve repeatedly noted) human persons are (or at least encompass) information processors somewhere in the Arithmetic Hierarchy, and (ii) (as we’ve also earlier noted) if persons are ordinary Turing machines they have a certain fixed size k, human persons are hypercomputers. QED\n\n---\n\nTitle: Available\nURL: http://kryten.mm.rpi.edu/sb_etal_godel_bb.pdf\nPublished: N/A\nAuthor: N/A\nHighlights:\nTuring-uncomputable\n...\n““busy\n...\nbeaver”\n...\nhypercompute.\n...\nuncomputable\n...\nbeaver”\n\n---\n\nTitle: An Overview of Independence Results of Set Theory and Open Mathematical Problems in Computational Complexity\nURL: https://www.thi.uni-hannover.de/fileadmin/thi/abschlussarbeiten/2021/Webseite_An_Overview_of_Independence_Results_of_Set_Theory_and_Open_Mathematical_Problems_in_Computational_Complexity.pdf\nPublished: N/A\nAuthor: N/A\nHighlights:\nA quite alluring part of theoretical computer science is dealing with the existence of non-computable functions. In other words, problems for which there are no terminating algorithms. Thanks to Church’s thesis, we can safely affirm that being computable is equivalent to being Turing-computable. Turing machines are the universal model of computation as they can translate any computing the human brain could perform. In this regard, a famous non-Turing-computable problem that has eluded many mathematicians is the function devised by Tibor Radó, known as the Busy Beaver function in the modern era. The latter is exceptionally rapidly growing so that it even exceeds the fast-growing Ackermann function. An interesting question regarding Busy Beaver is to determine the smallest value of n for which BB(n) is independent of the modern axioms of mathematics like ZFC . As an opening, we present an introduction to Busy Beaver and set theory. In the second chapter, we describe the programming language named Laconic developed by Adam Yedidia, with which he determines an upper bound to the question raised above. We also cite an overview of another related language serving the same purpose and producing enhanced results. Furthermore, we use Laconic to implement a few open mathematical problems both in number theory and graph theory, thereby proving the undecidability of several Busy Beaver values. Lastly, we tackle the famous and strenuous P vs. NP problem by demonstrating the adversity of proving its independence from ZF set theory.\n...\nThe delta transition function uniquely determines the behavior of the Turing machine. It takes two inputs: the current state and the symbol read from the tape. It outputs the next state, the symbol to be written on the tape, and the next direction of the head. Semantically, a Busy Beaver Turing machine with n states is the machine that produces the maximum number of transitions among all other n-state Turing machines before halting, knowing that it will eventually halt. The Busy Beaver function, therefore, BB(n) or the shift function S(n), as defined by Radó [RAD61], computes the number of transitions it takes a n-state Busy Beaver to halt. Radó also described the function Σ(n), which outputs the number of 1s (here bs) written on the tape before the machine halts. However, in this paper we stick with BB(n) as it is more intuitive.\n...\nδ transition function could have. \nLet M be a Turing machine that terminates. Let s(M) be the number of steps or δ \ntransitions that M takes before halting. If M stops at the first step, then s(M) = 1, \nand if M loops endlessly, we set s(M) = ∞ . \nAt this stage, we can formally define our Busy Beaver function BB(n) (or S(n)) as \nthe following: \n\nBB(n) = S(n) := max M ∈ T(n) | s(M) < ∞\n...\ns(M)\n...\nInformally, we range over all the finitely many n-state Turing machines. If we can safely \naffirm that a Turing machine runs forever, then we disregard it as this contradicts the \ndefinition of a Busy Beaver. Afterward, we maximize over the number of steps the \nhalting Turing machines take before eventually transitioning into the HALT state. \nIn his publication [RAD61], Radó describes this iterating over n-state halting Turing \nmachines as the Busy Beaver game. The latter represents a competition, in which \nthe winners are machines that achieve the maximum number of steps before halting, \nthereby determining the corresponding value of BB(n). \nPrior to presenting a straightforward proof of the non-computability of BB in \nconjunction with the famous halting problem, let’s start by defining the terms \ncomputability and decidability. The following definitions are present in [Vol20]. \nDefinition 1.4 A function is computabe iff it is Turing-computable. \nA function f : Σ∗ → ∆∗ is Turing-computable, if there is a deterministic Turing \nmachine M, so that for all x ∈ Σ∗ and y ∈ ∆∗: f(x) = y =⇒ M on input x halts with \n. . . y . . . on the output tape (in case of\n...\n, given a natural k as input, outputs BB(k) on \nits output tape. We prove that the characteristic function of the halting problem is \ncomputable: \nAlgorithm 1 Proof of decidability of the halting problem \nInput: w#x \n1: create Mw with w being a proper godelization of a Turing machine. Let n be the \nnumber of its states \n2: compute BB(n) \n3: run Mw on input x for BB(n) steps \n4: If Mw has not halted yet, then we can evidently conclude that Mw will never halt, \nbecause BB(n) is the maximum number of steps a n-state halting machine can \nperform before halting. Output 0 \n5: if Mw halted before or at BB(n) steps then output 1\n...\nThus, we have just proved that the halting problem is decidable, which we know is impossible. As a result, the Busy Beaver function is not computable. Historically, the Busy Beaver game has attracted considerable interest in the quest of determining as many BB(n) values as possible. The highest lower bound on the provable Busy Beaver value in modern mathematics systems is 4. Nonetheless, lower bounds have been proposed for 5, 6 and 7-state Busy Beavers:\n...\nNaturally, one could raise an interesting question on the computability of Busy Beaver values, namely, if we are able to prove BB(n) for n ∈ { 1, . . . , 4} , then why don’t we just utilize the same computation process of ranging overall potential Busy Beavers in order to determine BB(n) for n ≥ 4. The reason why this is not quite feasible is the existence of Turing machines, whose behavior is not decisively ascertainable. An example is the Turing machine M that we described in the previous chapter. Let n be the number of states of M. As the decision of whether M halts or runs forever is independent of ZFC axiomatic set theory, then the value of BB(n) is subsequently unprovable in ZFC . Suppose that the latter is provable, then ZFC can easily prove the behavior of M. Such proof will consist of running M and keeping track of the entirety of its transitions. If it halts, then the memorized computation history of the machine is the halting proof. If it doesn’t, then after exceeding the threshold of BB(n) steps and by definition of BB, we have proved that M will loop endlessly. Therefore, the machine M (also called Gödel Machine) suggests another perspective of approaching the Busy Beaver contest. The question of proving all Busy Beaver values becomes the question of how many of them are knowable using a fixed axiomatic arithmetic-encoding theory as a basis for our computations. In other words, from which n the value BB(n) eludes modern mathematics. Adam Yedidia, Scott Aaronson, and Stefan O’rear considered this dilemma as a powerful motivation for their works of finding the highest upper bound, for which the Busy Beaver value is provable in ZFC set theory. [YA16] Theorem 1.1 If T is a computable and arithmetic-encoding consistent axiomatic theory. Then, there exists a constant nT ∈ N, so that for all n ≥ nT , the value of BB(n) cannot be proved in T. Proof. Let M be our Turing machine discussed above that, on the all-blank symbol input, enumerates the axioms and provable theorems and conclusions of T and terminates iff it finds a contradiction. However, since we assumed that T is consistent, we know in advance that M will never halt. Nevertheless, T cannot prove that M will never halt. If it could, then we have ZF | = Con(ZF), which is a direct violation of the second Gödel’s incompleteness theorem. Now let nT be the number of states in M, then, for all n ≥ nT , BB(n) is unprovable in T. If such proof existed, then T will be able to prove that M never halts, simply by simulation M for BB(n) steps. If M hasn’t halted yet, then we know it will never halt because BB(n) ≥ BB(nT ). In the next chapter, we will be discussing, among other topics, the achievements reached by Yedidia, Aaronson, and O’rear and the ideas implemented in the goal of finding the smallest possible n for which BB(n) is independent of ZFC set theory.\n\n---\n\nTitle: New Sequence Problems and Solutions from 2022\nURL: https://sites.math.rutgers.edu/~zeilberg/expmath/sloane22.pdf\nPublished: N/A\nAuthor: N/A\nHighlights:\nIs it Augusto Santi’s 2269429312765395470820 in A351871 ? Is it Pavel Kropitz’s new lower bound on Busy Beaver(6) ?\n...\nA060843: BB(1) = 1, BB(2) = 6, BB(3) = 21, BB(4) = 107, BB(5) = ?, BB(6) = ?\n...\nBB(6) > 10^^15 (A tower of 10’s of height 15)\n...\nNo! 10^^log n from Kaprekar’s Junction Numbers (see below)\n\n---\n\nTitle: Noncomputable Processes - Bibliography - PhilPapers\nURL: https://philpapers.org/browse/noncomputable-processes\nPublished: N/A\nAuthor: N/A\nHighlights:\n| Summary | In computability theory the class of functions that are computable corresponds to the class of recursive ones. This was proven by Church's lambda-calculus by formulating the former as lambda terms. Turing proved that this class also coincides with the class of functions that are mechanically computable. These results are usually combined in the so-called Church-Turing Thesis: every effectively calculable function is a computable function. This statement remains a thesis (and it thus not a theorem) because there is no formal evidence that non-computable processes are besides the limits of (mechanical or formal) calculability. Nonetheless, it is usually considered true on account of the large amount of evidence in its favour. This immediately defines the class of non-computable function as those that are non-recursively definable and the class of non-solvable problems by means of recursive procedures. Among the most common examples of non-computable function is the busy-beaver function (also known as the productivity function): consider the function executed by a Turing machine that starts on a blank tape and when it halts after scanning a block of strokes, the length of that block is said to be its productivity; but if it does not halt, then its productivity is 0; then the busy-beaver function p(n) is the productivity of the most productive Turing machine of a given class, i.e. with at most n states. Such function corresponds to the Halting Function, based on the idea that all Turing machines can be enumerated and a general procedure can be defined to establish for any given machine and any given input whether that machine with that input halts or does not. This moreover coincides with the Decision Problem for First Order Logic: given a finite set of sentences S, for any given sentence D, establish whether S implies D (in FOL). From a philosophical viewpoint, the uncomputable has generated an enormous amount of work, for example concerning the physical non-computability of non-recursive function (on the assumption that the universe behaves as a Turing Machine); or concerning the limits of Turing-computable mental states. The thesis that it does exist a class of computable functions beyond Turing computability is usually referred to as 'hypercomputability', or 'super-Turing computability' and the supposed corresponding class of algorithms is known as 'super-recursive'. |\n...\nof Computing and Information\n...\nNoncomputable Processes\n...\nPhilosophy of Computing and Information\n...\nNoncomputable Processes\n...\nPhilosophy of Computing\n...\nNoncomputable Processes in Philosophy of Computing and Information\n...\nNoncomputable Processes\n...\nComputing and Information", "metadata": { "provider": "exa", "truncated": false }, "title": "Exa Web Search: \"Busy Beaver\" \"entry threshold\" Codeforces 2257E", "time": { "start": 1787510155169, "end": 1787510156584 } }, "metadata": { "openai": { "itemId": "fc_0492323e0977dd17016a8b3d8afe3087d0afeed397d359e934" } }, "id": "prt_02fe867a0001VWJ7C9a6NfUi3q", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe77de4001LPI6Rn6TXaN2mm" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 22193, "input": 3145, "output": 232, "reasoning": 2048, "cache": { "write": 0, "read": 16768 } }, "cost": 0, "id": "prt_02fe86d2d001ulp48TWZIY2Dil", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe77de4001LPI6Rn6TXaN2mm" } ] }, { "info": { "parentID": "msg_02fe71263001VNs7XA6YAFu76N", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "max", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-I0zj28", "root": "/" }, "cost": 0, "tokens": { "total": 66879, "input": 36243, "output": 3162, "reasoning": 8658, "cache": { "write": 0, "read": 18816 } }, "modelID": "gpt-5.6-sol", "providerID": "openai", "time": { "created": 1787510156600, "completed": 1787510441599 }, "finish": "tool-calls", "structured": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "For one floor, the budget changes by $d_{i,j}=b_{i,j}-a_{i,j}$, but you still need at least $a_{i,j}$ before starting it. Try to separate the problem into two phases: first get as many carrots as possible, then spend them to push one chosen building as high as possible." }, { "order": 2, "content": "In one building, start at its next unbuilt floor and take the shortest prefix whose total net change is **strictly positive**. Call it an earning block. Every proper prefix of this block has non-positive net change, so stopping halfway cannot help finance work elsewhere. If no such endpoint exists, every prefix of the remaining suffix is non-positive." }, { "order": 3, "content": "Suppose a block starts with capital $C$. If $s$ is its net change before floor $j$, that floor requires $C+s\\ge a_j$. Therefore the whole block has entry requirement\n$$R=\\max_j(a_j-s_j),$$\nwhere $s_j$ is the net change before floor $j$. Its gain is $P=\\sum_j(b_j-a_j)>0$. Scan each building once to compute all triples $(R,P,\\text{ending floor})$." }, { "order": 4, "content": "Only the first unfinished earning block of each building is currently available. Put these blocks in a min-heap by $R$. While the minimum $R$ is at most the current capital, complete that block, add its positive gain, and insert the next block from the same building. The heap order is not some mystical ROI trick: gains never hurt, and the minimum tells you whether **any** available block is affordable." }, { "order": 5, "content": "When the minimum requirement exceeds the current capital $C$, no strategy can earn more. Before the first remaining earning block could finish, every building would only contribute a non-positive proper-prefix change, so interleaving cannot help meet that block's requirement. Now, for every building independently, start with $C$ and simulate its remaining floors until the next cost is unaffordable. Work on other remaining buildings could only subtract a non-negative amount from your budget, so this direct simulation is optimal. Take the largest height and then the smallest index." } ], "editorial": "Let the net change from floor $j$ of building $i$ be\n$$\nd_{i,j}=b_{i,j}-a_{i,j}.\n$$\nAfter constructing that floor, the budget changes by $d_{i,j}$. However, before construction, the budget must still be at least $a_{i,j}$. That upfront requirement is the whole annoyance here.\n\nA DP over the budget is dead on arrival because $x$ can be $10^{18}$. Instead, we compress each project into useful earning opportunities.\n\n## Earning blocks\n\nConsider one building and its next unbuilt floor. Scan upward until the running sum of $d$ becomes **strictly positive** for the first time. These floors form one earning block. Then start another block at the next floor and repeat.\n\nIf a block covers floors $l,\\ldots,r$, then\n$$\nP=\\sum_{j=l}^{r}d_j>0,\n$$\nwhile every proper prefix satisfies\n$$\n\\sum_{j=l}^{k}d_j\\le 0 \\qquad (l\\le k0$: its net gain;\n- $r$: the height after completing it.\n\nWe compute them during the same scan. Keep `balance`, the current $s_j$, and update\n$$\nR\\leftarrow\\max(R,a_j-\\text{balance}),\n$$\nthen add $b_j-a_j$ to `balance`. When `balance` becomes positive, save the block and reset both values.\n\n## Farming all available money\n\nFor each building, only its first unfinished block is available because higher floors cannot be skipped. Put one available block per building into a min-heap ordered by $R$.\n\nLet the current capital be $C$.\n\n- If the heap minimum has $R\\le C$, complete it.\n- Increase $C$ by its gain $P$.\n- Record its ending floor.\n- Insert the next block of the same building, if one exists.\n- If the minimum has $R>C$, stop.\n\nCompleting an affordable block cannot hurt: its final gain is positive, it advances one building, and it cannot make any other contract less affordable. We do not choose the minimum because it has a better profit ratio or any other finance-bro nonsense. The minimum is useful because, if even it is too expensive, every currently available block is too expensive.\n\n## Why stopping really means stopping\n\nSuppose the heap has stopped at capital $C$, and imagine that some continuation manages to complete another earning block. Look at the **first** remaining earning block completed by that continuation.\n\nBefore this first completion, each building has constructed only a proper prefix of its current block, or a prefix of its final suffix. By construction, each such prefix has net change at most zero. Therefore all work done before that completion can only keep the capital at or below $C$.\n\nThe block being completed was one of the currently available blocks; reaching a later block would require completing an earlier one first. Its requirement is greater than $C`, because the heap stopped. Interleaving other non-profitable prefixes only gives it less money, not more. So completing it is impossible, a contradiction.\n\nHence no remaining work can increase the capital beyond $C$. This is the maximum amount of money we can gather before making the final push for height.\n\n## Building the final tower\n\nLet `built[i]` be the height reached in building $i$ by completed earning blocks. For each candidate building $i$ independently:\n\n1. Start with the maximum capital $C$.\n2. Start at floor `built[i]`.\n3. While the current capital is at least the next $a_{i,j}$, construct that floor and apply $b_{i,j}-a_{i,j}$.\n4. Record the resulting height.\n\nWhy can we ignore unfinished floors of other buildings now? After the earning phase, no positive block can be completed. At every moment, the remaining work done in any other building is only a proper block prefix or a final-suffix prefix, whose total net change is at most zero. If we have already built $k$ extra floors of the target, working elsewhere can only leave us with no more money than the target-only simulation has after those same $k$ floors. Thus it cannot make the next target floor affordable.\n\nSo the independent simulation gives the maximum possible height for each building. Scan buildings in increasing index order and update the answer only for a strictly larger height; ties then automatically keep the smallest index.\n\n## Correctness proof\n\n**Lemma 1.** A block with entry requirement $R$ can be completed from capital $C$ if and only if $C\\ge R$.\n\n**Proof.** Before each floor $j$, the available capital is $C+s_j$. That floor is affordable exactly when $C\\ge a_j-s_j$. All floors are affordable exactly when $C$ is at least the maximum of these values, which is $R$. $\\square$\n\n**Lemma 2.** Completing any available block with $R\\le C$ cannot reduce the set of future possibilities.\n\n**Proof.** By Lemma 1, the block can be completed. Its gain $P$ is positive, so afterward the capital is $C+P>C$. No other building changes, and this building only unlocks its next floors. Therefore every action that was possible before remains possible afterward. $\\square$\n\n**Lemma 3.** When the heap process stops, its capital $C$ is the maximum reachable capital.\n\n**Proof.** By Lemma 2, every block chosen by the heap is safe to perform before any non-earning work. At termination, every available block has requirement greater than $C$. If a continuation completed some earning block, consider the first one it completed. Before that event, every constructed remainder in every building is a proper block prefix or a final-suffix prefix, so each contributes net change at most zero. The continuation never has more than $C$ before completing the block. Since that available block requires more than $C$, Lemma 1 says it cannot be completed. This contradiction shows that no continuation can gain more money. $\\square$\n\n**Lemma 4.** From the terminal state, simulating only the remaining floors of a chosen building constructs the maximum possible number of its floors.\n\n**Proof.** After $k$ additional target floors, the target-only simulation has $C$ plus their net change. Any other work that can occur before the next target floor is a collection of non-positive prefixes, by Lemma 3 and the block construction. It cannot increase that amount. Therefore, if the target-only simulation cannot afford the next floor, no interleaved schedule can afford it either. $\\square$\n\n**Theorem.** The algorithm outputs the maximum achievable building height and the smallest building index achieving it.\n\n**Proof.** Lemma 3 gives the largest useful starting capital for the final phase. Lemma 4 computes the exact maximum height of every building from that state. Taking the largest of these heights is optimal, and processing indices in increasing order resolves ties toward the smallest index. $\\square$\n\n## Complexity\n\nLet $M=\\sum_i m_i$ for one test case. Every floor is scanned once while forming blocks and at most once during the final simulations. Every block enters and leaves a heap of size at most $n$ once.\n\n$$\n\\text{Time}=O(M\\log n),\\qquad \\text{Memory}=O(M).\n$$\n\nAll budget, gain, and requirement values must use 64-bit integers.", "solution": "#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nstruct Block {\n ll need, gain;\n int end;\n};\n\nint main() {\n setIO();\n\n int t;\n cin >> t;\n while (t--) {\n int n;\n ll x;\n cin >> n >> x;\n\n vector> a(n), b(n);\n vector> blocks(n);\n\n for (int i = 0; i < n; ++i) {\n int m;\n cin >> m;\n a[i].resize(m);\n b[i].resize(m);\n for (ll &v : a[i]) cin >> v;\n for (ll &v : b[i]) cin >> v;\n\n ll balance = 0, need = 0;\n for (int j = 0; j < m; ++j) {\n need = max(need, a[i][j] - balance);\n balance += b[i][j] - a[i][j];\n if (balance > 0) {\n blocks[i].push_back({need, balance, j + 1});\n balance = need = 0;\n }\n }\n }\n\n priority_queue,\n vector>,\n greater>> pq;\n vector nextBlock(n, 0), built(n, 0);\n\n for (int i = 0; i < n; ++i) {\n if (!blocks[i].empty()) {\n pq.emplace(blocks[i][0].need, i);\n }\n }\n\n while (!pq.empty() && pq.top().first <= x) {\n auto [required, i] = pq.top();\n pq.pop();\n\n const Block &block = blocks[i][nextBlock[i]++];\n x += block.gain;\n built[i] = block.end;\n\n if (nextBlock[i] < (int)blocks[i].size()) {\n pq.emplace(blocks[i][nextBlock[i]].need, i);\n }\n }\n\n int bestHeight = -1, answer = 0;\n for (int i = 0; i < n; ++i) {\n ll money = x;\n int floor = built[i];\n while (floor < (int)a[i].size() && money >= a[i][floor]) {\n money += b[i][floor] - a[i][floor];\n ++floor;\n }\n if (floor > bestHeight) {\n bestHeight = floor;\n answer = i;\n }\n }\n\n cout << bestHeight << ' ' << answer + 1 << '\\n';\n }\n\n return 0;\n}\n" }, "id": "msg_02fe86d38001O3R9xjdKI1mpwO", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb" }, "parts": [ { "type": "step-start", "id": "prt_02fe872bc001aAwrESoK5QM59g", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" }, { "type": "reasoning", "text": "**Evaluating zero-net segment handling**", "time": { "start": 1787510158016, "end": 1787510168151 }, "metadata": { "openai": { "itemId": "rs_0a754394792919ad016a8b3d8df1fc87d0998a57b8d9c78fb2" } }, "id": "prt_02fe872c0001MDlix7RZlK5MeD", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" }, { "type": "reasoning", "text": "**Confirming zero segment merging validity**", "time": { "start": 1787510168153, "end": 1787510171850 }, "metadata": { "openai": { "itemId": "rs_0a754394792919ad016a8b3d8df1fc87d0998a57b8d9c78fb2" } }, "id": "prt_02fe89a59001hkM9Jrh5A65N46", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" }, { "type": "reasoning", "text": "**Validating segment threshold formula**", "time": { "start": 1787510171853, "end": 1787510171881 }, "metadata": { "openai": { "itemId": 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} }, "id": "prt_02febf2510011L1la81FK8rHj3", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" }, { "type": "tool", "tool": "StructuredOutput", "callID": "call_TvSc1w8O4fG0NrcCKfMkXyL1", "state": { "status": "completed", "input": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "For one floor, the budget changes by $d_{i,j}=b_{i,j}-a_{i,j}$, but you still need at least $a_{i,j}$ before starting it. Try to separate the problem into two phases: first get as many carrots as possible, then spend them to push one chosen building as high as possible." }, { "order": 2, "content": "In one building, start at its next unbuilt floor and take the shortest prefix whose total net change is **strictly positive**. Call it an earning block. Every proper prefix of this block has non-positive net change, so stopping halfway cannot help finance work elsewhere. If no such endpoint exists, every prefix of the remaining suffix is non-positive." }, { "order": 3, "content": "Suppose a block starts with capital $C$. If $s$ is its net change before floor $j$, that floor requires $C+s\\ge a_j$. Therefore the whole block has entry requirement\n$$R=\\max_j(a_j-s_j),$$\nwhere $s_j$ is the net change before floor $j$. Its gain is $P=\\sum_j(b_j-a_j)>0$. Scan each building once to compute all triples $(R,P,\\text{ending floor})$." }, { "order": 4, "content": "Only the first unfinished earning block of each building is currently available. Put these blocks in a min-heap by $R$. While the minimum $R$ is at most the current capital, complete that block, add its positive gain, and insert the next block from the same building. The heap order is not some mystical ROI trick: gains never hurt, and the minimum tells you whether **any** available block is affordable." }, { "order": 5, "content": "When the minimum requirement exceeds the current capital $C$, no strategy can earn more. Before the first remaining earning block could finish, every building would only contribute a non-positive proper-prefix change, so interleaving cannot help meet that block's requirement. Now, for every building independently, start with $C$ and simulate its remaining floors until the next cost is unaffordable. Work on other remaining buildings could only subtract a non-negative amount from your budget, so this direct simulation is optimal. Take the largest height and then the smallest index." } ], "editorial": "Let the net change from floor $j$ of building $i$ be\n$$\nd_{i,j}=b_{i,j}-a_{i,j}.\n$$\nAfter constructing that floor, the budget changes by $d_{i,j}$. However, before construction, the budget must still be at least $a_{i,j}$. That upfront requirement is the whole annoyance here.\n\nA DP over the budget is dead on arrival because $x$ can be $10^{18}$. Instead, we compress each project into useful earning opportunities.\n\n## Earning blocks\n\nConsider one building and its next unbuilt floor. Scan upward until the running sum of $d$ becomes **strictly positive** for the first time. These floors form one earning block. Then start another block at the next floor and repeat.\n\nIf a block covers floors $l,\\ldots,r$, then\n$$\nP=\\sum_{j=l}^{r}d_j>0,\n$$\nwhile every proper prefix satisfies\n$$\n\\sum_{j=l}^{k}d_j\\le 0 \\qquad (l\\le k0$: its net gain;\n- $r$: the height after completing it.\n\nWe compute them during the same scan. Keep `balance`, the current $s_j$, and update\n$$\nR\\leftarrow\\max(R,a_j-\\text{balance}),\n$$\nthen add $b_j-a_j$ to `balance`. When `balance` becomes positive, save the block and reset both values.\n\n## Farming all available money\n\nFor each building, only its first unfinished block is available because higher floors cannot be skipped. Put one available block per building into a min-heap ordered by $R$.\n\nLet the current capital be $C$.\n\n- If the heap minimum has $R\\le C$, complete it.\n- Increase $C$ by its gain $P$.\n- Record its ending floor.\n- Insert the next block of the same building, if one exists.\n- If the minimum has $R>C$, stop.\n\nCompleting an affordable block cannot hurt: its final gain is positive, it advances one building, and it cannot make any other contract less affordable. We do not choose the minimum because it has a better profit ratio or any other finance-bro nonsense. The minimum is useful because, if even it is too expensive, every currently available block is too expensive.\n\n## Why stopping really means stopping\n\nSuppose the heap has stopped at capital $C$, and imagine that some continuation manages to complete another earning block. Look at the **first** remaining earning block completed by that continuation.\n\nBefore this first completion, each building has constructed only a proper prefix of its current block, or a prefix of its final suffix. By construction, each such prefix has net change at most zero. Therefore all work done before that completion can only keep the capital at or below $C$.\n\nThe block being completed was one of the currently available blocks; reaching a later block would require completing an earlier one first. Its requirement is greater than $C`, because the heap stopped. Interleaving other non-profitable prefixes only gives it less money, not more. So completing it is impossible, a contradiction.\n\nHence no remaining work can increase the capital beyond $C$. This is the maximum amount of money we can gather before making the final push for height.\n\n## Building the final tower\n\nLet `built[i]` be the height reached in building $i$ by completed earning blocks. For each candidate building $i$ independently:\n\n1. Start with the maximum capital $C$.\n2. Start at floor `built[i]`.\n3. While the current capital is at least the next $a_{i,j}$, construct that floor and apply $b_{i,j}-a_{i,j}$.\n4. Record the resulting height.\n\nWhy can we ignore unfinished floors of other buildings now? After the earning phase, no positive block can be completed. At every moment, the remaining work done in any other building is only a proper block prefix or a final-suffix prefix, whose total net change is at most zero. If we have already built $k$ extra floors of the target, working elsewhere can only leave us with no more money than the target-only simulation has after those same $k$ floors. Thus it cannot make the next target floor affordable.\n\nSo the independent simulation gives the maximum possible height for each building. Scan buildings in increasing index order and update the answer only for a strictly larger height; ties then automatically keep the smallest index.\n\n## Correctness proof\n\n**Lemma 1.** A block with entry requirement $R$ can be completed from capital $C$ if and only if $C\\ge R$.\n\n**Proof.** Before each floor $j$, the available capital is $C+s_j$. That floor is affordable exactly when $C\\ge a_j-s_j$. All floors are affordable exactly when $C$ is at least the maximum of these values, which is $R$. $\\square$\n\n**Lemma 2.** Completing any available block with $R\\le C$ cannot reduce the set of future possibilities.\n\n**Proof.** By Lemma 1, the block can be completed. Its gain $P$ is positive, so afterward the capital is $C+P>C$. No other building changes, and this building only unlocks its next floors. Therefore every action that was possible before remains possible afterward. $\\square$\n\n**Lemma 3.** When the heap process stops, its capital $C$ is the maximum reachable capital.\n\n**Proof.** By Lemma 2, every block chosen by the heap is safe to perform before any non-earning work. At termination, every available block has requirement greater than $C$. If a continuation completed some earning block, consider the first one it completed. Before that event, every constructed remainder in every building is a proper block prefix or a final-suffix prefix, so each contributes net change at most zero. The continuation never has more than $C$ before completing the block. Since that available block requires more than $C$, Lemma 1 says it cannot be completed. This contradiction shows that no continuation can gain more money. $\\square$\n\n**Lemma 4.** From the terminal state, simulating only the remaining floors of a chosen building constructs the maximum possible number of its floors.\n\n**Proof.** After $k$ additional target floors, the target-only simulation has $C$ plus their net change. Any other work that can occur before the next target floor is a collection of non-positive prefixes, by Lemma 3 and the block construction. It cannot increase that amount. Therefore, if the target-only simulation cannot afford the next floor, no interleaved schedule can afford it either. $\\square$\n\n**Theorem.** The algorithm outputs the maximum achievable building height and the smallest building index achieving it.\n\n**Proof.** Lemma 3 gives the largest useful starting capital for the final phase. Lemma 4 computes the exact maximum height of every building from that state. Taking the largest of these heights is optimal, and processing indices in increasing order resolves ties toward the smallest index. $\\square$\n\n## Complexity\n\nLet $M=\\sum_i m_i$ for one test case. Every floor is scanned once while forming blocks and at most once during the final simulations. Every block enters and leaves a heap of size at most $n$ once.\n\n$$\n\\text{Time}=O(M\\log n),\\qquad \\text{Memory}=O(M).\n$$\n\nAll budget, gain, and requirement values must use 64-bit integers.", "solution": "#include \nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n}\n\nstruct Block {\n ll need, gain;\n int end;\n};\n\nint main() {\n setIO();\n\n int t;\n cin >> t;\n while (t--) {\n int n;\n ll x;\n cin >> n >> x;\n\n vector> a(n), b(n);\n vector> blocks(n);\n\n for (int i = 0; i < n; ++i) {\n int m;\n cin >> m;\n a[i].resize(m);\n b[i].resize(m);\n for (ll &v : a[i]) cin >> v;\n for (ll &v : b[i]) cin >> v;\n\n ll balance = 0, need = 0;\n for (int j = 0; j < m; ++j) {\n need = max(need, a[i][j] - balance);\n balance += b[i][j] - a[i][j];\n if (balance > 0) {\n blocks[i].push_back({need, balance, j + 1});\n balance = need = 0;\n }\n }\n }\n\n priority_queue,\n vector>,\n greater>> pq;\n vector nextBlock(n, 0), built(n, 0);\n\n for (int i = 0; i < n; ++i) {\n if (!blocks[i].empty()) {\n pq.emplace(blocks[i][0].need, i);\n }\n }\n\n while (!pq.empty() && pq.top().first <= x) {\n auto [required, i] = pq.top();\n pq.pop();\n\n const Block &block = blocks[i][nextBlock[i]++];\n x += block.gain;\n built[i] = block.end;\n\n if (nextBlock[i] < (int)blocks[i].size()) {\n pq.emplace(blocks[i][nextBlock[i]].need, i);\n }\n }\n\n int bestHeight = -1, answer = 0;\n for (int i = 0; i < n; ++i) {\n ll money = x;\n int floor = built[i];\n while (floor < (int)a[i].size() && money >= a[i][floor]) {\n money += b[i][floor] - a[i][floor];\n ++floor;\n }\n if (floor > bestHeight) {\n bestHeight = floor;\n answer = i;\n }\n }\n\n cout << bestHeight << ' ' << answer + 1 << '\\n';\n }\n\n return 0;\n}\n" }, "output": "Structured output captured successfully.", "metadata": { "valid": true }, "title": "Structured Output", "time": { "start": 1787510441491, "end": 1787510441495 } }, "metadata": { "openai": { "itemId": "fc_0a754394792919ad016a8b3e734cd487d09a4a70aca45d9ae1" } }, "id": "prt_02febf2c4001RNS4ERTiUAy56q", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 66879, "input": 36243, "output": 3162, "reasoning": 8658, "cache": { "write": 0, "read": 18816 } }, "cost": 0, "id": "prt_02fecc67b001VKYf3HrPBoZLj6", "sessionID": "ses_fd018edaeffevjvXwJtcL2L9sb", "messageID": "msg_02fe86d38001O3R9xjdKI1mpwO" } ] } ] }