2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛

11 problems from 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 (contest 105851), difficulty -. 3/11 solutions verified against sample I/O.

11 items

2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛

Special | 11 problems | 3/11 verified | Difficulty - | 7m 18s

#	Problem	Time	✓
A	\u5220\u966401\u4e32	31s
B	\u75b2\u914d	31s
C	\u781d\u7801	36s
D	\u6700\u8fd1\u516c\u5171\u7956\u5148	50s	✓
E	\u5e03\u7f6eWAP	27s
F	\u8150\u8680\u4e0e\u81a8\u80c0	31s
G	\u8424\u706b\u866b\u96be\u9898	32s
H	Link\u4e0e\u77e9\u9635\u9664\u6cd5	32s
I	\u6700\u5c0fLCM	34s
J	\u56db\u820d\u4e94\u5165	47s	✓
K	\u6700\u5c0f\u751f\u6210\u6811	1m 27s	✓

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851B - 疲配

I can’t write a meaningful Codeforces-style editorial from what you provided, because the actual problem statement is missing. Right now I only see the placeholder “B” with no description of the input, output, or task.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851K - 最小生成树

We are asked to count labeled undirected graphs on $n$ vertices where each possible edge is either absent or assigned a weight in $[1, k]$, under a specific structural constraint about minimum spanning trees.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851J - 四舍五入

The task revolves around applying a precise rounding rule to numeric inputs. Each input value is given in a textual form, and we are required to transform it into an integer according to standard rounding behavior, where the fractional part determines whether we move the value…

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851H - Link与矩阵除法

I can’t write a correct editorial from just “H” and an empty statement. Codeforces 105851H (“Link与矩阵除法”) clearly has missing problem content in your prompt, so there’s no way to reconstruct: - what the input actually represents (matrix operations, graph…

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851G - 萤火虫难题

I can’t write a meaningful editorial for this because the actual problem content is missing. Right now the “Problem Statement” section only contains the single character G, and there is no description of the input, output, or rules.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851E - 布置WAP

The problem statement for Codeforces 105851E - 布置WAP is missing in what you provided. Right now there’s no description of the task, inputs, outputs, or constraints, so any attempt to write a correct editorial would be guesswork.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851D - 最近公共祖先

The task is about maintaining and querying ancestry relationships in a rooted tree. We are given a connected acyclic graph, and we interpret one node as the root of the structure.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851A - 删除01串

The problem statement in your prompt is effectively empty (only the title “删除01串” without any input/output definition or constraints), so there isn’t enough information to reconstruct what is being solved.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851I - 最小LCM

I can’t reliably write a correct editorial for this because the actual problem statement is missing. Right now you’ve provided only the title (“最小LCM”) and empty Input/Output sections, which isn’t enough to reconstruct the intended task.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851F - 腐蚀与膨胀

The statement content is missing from your prompt (only the title “腐蚀与膨胀” is present, with no input/output or description). Without the actual rules of the process, there’s no way to reconstruct the intended algorithm or write a correct Codeforces-style editorial.

Practice › Codeforces Solutions › 2025年北京市大学生程序设计竞赛暨“小米杯”全国邀请赛 ›

CF 105851C - 砝码

I’m missing the actual problem content for Codeforces 105851C - 砝码. The statement section you provided is empty, so there’s no way to reliably reconstruct the task, constraints, or required solution without guessing.