AtCoder Beginner Contest 358 A - Welcome to AtCoder Land (abc358 A) 题目大意 给定两个字符串,问是否是AtCoder Land 解题思路 读取后判断即可。 神奇的代码 #include<bits/stdc++.h> usingnamespacestd; usingLL =longlong; intmain(void){ ios::
AtCoder Beginner Contest 358 合集- 记录随手刷过的编程题(59) 1.E. Arranging The Sheep2024-02-282.C. Team2024-02-293.C. RationalLee2024-03-014.B. Preparing Olympiad2024-03-055.B. Find The Array2024-03-056.E. Accidental Victory2024-03-067.A. Learning Languages2024-03-078.A. k-th ...
We will hold AtCoder Beginner Contest 358. Contest URL: https://atcoder.jp/contests/abc358 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240615T2100&p1=248 Duration: 100 minutes Writer: yuto1115, cn449 Tester: sotanishy, MtSaka Rated range: ~ 1999 The point v...
void solve() { int n, a; std::cin >> n >> a; std::vector<int> t(n); for (int i = 0; i < n; i++) { std::cin >> t[i]; } std::vector<int> ans(n + 1); for (int i = 1; i <= n; i++) { ans[i] = std::max(t[i - 1] + a, ans[i - 1] + a...
void solve() { int k; cin >> k; vector<int> a(30); for (int i = 1; i <= 26; ++i) cin >> a[i]; for (int i = 0; i <= 26; ++i) f[i][0] = 1; ll ans = 0; for (int i = 1; i <= k; ++i) { for (int j = 1; j <= 26; ++j) { f[j][i] =...
We will hold AtCoder Beginner Contest 358. Contest URL:https://atcoder.jp/contests/abc358 Start Time:http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240615T2100&p1=248 Duration: 100 minutes Writer:yuto1115,cn449 Tester:sotanishy,MtSaka ...
·AtCoder Beginner Contest (ABC) 这是最频繁且最简单的入门赛,通常情况下每月至少举行2次。2019年4月27日(含)之前,每场比赛共4题,时长100分钟,满分1000分且Rating超过1199的选手不计Rating值。自2019年5月19日起改版升级为6道题目,时长不变,满分2100分且Rating值超过1999的选手不计Rating值。改版之后比赛质量...
【AtCoder Beginner Contest 直播活动暂停通知】自 2024 年 7 月起,鉴于老师目前的工作负担,我们遗憾地宣布每周的 AtCoder Beginner Contest 直播讲题活动将暂停。请持续关注我们的粉专或 IG ,一旦活动恢复,我们将第一时间通知大家。比赛链接:https://atcoder.jp/contests/abc359 --- 01:22 A - 循环 02:55 ...
Σ 大赛——AtCoder Beginner Contest 353 https://www.bilibili.com/video/BV1kt421u7XK/ https://www.bilibili.com/video/BV1vf42127tH/ AtCoder Beginner Contest 353 实况(A~E) https://www.bilibili.com/video/BV1Gs421N748/ https://www.bilibili.com/video/BV19E42137MQ/ AtCoder Beginner Cont...
AtCoder Beginner Contest 174 题解 AtCoder Beginner Contest 174 题解C Repsept 知识点:取模对加法和乘法封闭,所以暴力到2×1062\times 10^62×106,一边取模一边特判即可。D - Alter Altar 最终肯定是RRR全在左边,简单证明:假设最终RRR左边的某个位置存在WWW,则该位置右边只能是WWW,然后就是子问题了...