Codeforces Round 1024 (Div. 1, Div. 2) ByShoo,3 weeks ago, Hello, Codeforces! eren__,sweetweasel, and I are glad to invite you toCodeforces Round 1024 (Div. 1)andCodeforces Round 1024 (Div. 2). Both rounds take place atMay/11/2025 17:35 (Moscow time), and have a duration of2...
Before contestCodeforces Round 1027 (Div. 3)43:40:15Register now » → Filter Problems Difficulty: — Add tag Main acmsguru | Problems Submit Status Standings Custom testProblems #Name 2110F Faculty math x287 2110E Melody dfs and similar, graphs x657 ...
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CodeForces Solution Codeforces is a website that hosts competitive programming contests. It is maintained by a group of competitive programmers from ITMO University led by Mikhail Mirzayanov. Since 2013, Codeforces claims to surpass Topcoder in terms of active contestants. As of 2018, it has over...
View Code B. Ehab and subtraction Water. View Code C. Ehab and a 2-operation task Solved. 题意: 有两种操作 将前j个数全都加上x将前j个数全都加上x 将前j个数全都modx将前j个数全都modx 要求用不超过n+1次操作,使得给定序列变成严格的上升序列n+1次操作,使得给定序列变成严格的上升序列 ...
给你五个数 n,a,b,p,q,有 n 个蛋糕编号为 1,2,\dots,n,将编号为 a 的倍数的蛋糕染成红色对答案产生 p 的贡献,同理将所有编号为 b 的倍数的蛋糕染成蓝色对答案产生 q 的贡献。若一个蛋糕的编号同时 a,b 的倍数,则可以自行选择颜色染色。求答案最大值。 简要分析 有重复的情况染贡献最大值所对应...
给你一个长度为 n 的权值数组与一个长度为 n 的字符串,仅由 A,B 组成。 A 表示Alice 会得到权值,B 表示Bob 会得到权值,可以翻转一次权值数组的前后缀,这里指 从 A→B。 问Bob 能获得的最大权值。 简要分析 预处理翻或不翻的前后缀,枚举即可。 代码实现 void solve() { ccin >> n; for (ll i...
Educational Codeforces Round 55 (Rated for Div. 2) Solution,A.VasyaandBookSolved.三种方式取$Min$1#include<bits/stdc++.h>2usingnamespacestd;34#definelllonglong5#defineINF0x3f3f3f3f3f3f3f3f6int
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequenceaand thecompressed sequenceof the sequencecwere known from that test.
Codeforces 543E. Listening to Music,Description"题面"Solution分块套分块,分别对时间和位置进行分块差不多是一个定期保存信息的方法对于询问我们不妨求出$=x$的答案,然后用$m(=x)$的答案,避免了严格小于带来的麻烦暴力做法是把数字从大到小加入,然后每一次做区间修改,主