A.First ABC 2模拟即可。 inlinevoidqfl_zzz(){ll n=read();string s=sread();for(ll i=1;i<=n-2;++i)if(s[i]=='A'&&s[i+1]=='B'&&s[i+2]=='C'){writen(i);return;}writen(-1);} B.Prefix and Suffix模拟即可。 inlinevoidqfl_zzz(){ll n=read(),m=read();string s=sr...
ABC 345 F - Many Lamps03-17 收起 AtCoder Beginner Contest 332A - Online Shopping代码:highlighter- cpp#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e6 + 10; typedef pair<ll, ll> pii; void solve() ...
AtCoder Beginner Contest 322 Cranewilliams 天青色等烟雨,而我在等你 A - First ABC 2 c++库函数substr B - Prefix and Suffix c++库函数substr(没错,你没看错,同上) C - Festival 倒着维护一个dp值 D - Polyomino 有意思的一道模拟题 方格4*4,给足了想象空间。 我的思路很 simple 就是,枚举三个图...
AtCoder Beginner Contest 322 合集- Atcoder 日常vp训练(5) 1.AtCoder ABC 2162023-08-052.AtCoder ABC 2182023-08-093.AtCoder Beginner Contest 2292023-09-20 4.AtCoder Beginner Contest 3222023-10-015.AtCoder Beginner Contest 1742024-04-12 收起 ...
AtCoder Beginner Contest 322 A 至 F 題讲解 by dreamoon https://www.bilibili.com/video/BV1hN411E7GL/ AtCoder Beginner Contest 323 A.Weak Beats讲解 https://www.bilibili.com/video/BV1HN4y117TH/ Atcoder ABC324A-F讲解 https://www.bilibili.com/video/BV1sh4y1q7wJ/ ...
0Tags Code Folders and files Name Last commit message Last commit date Latest commit yukikotani231 Create README.md Oct 10, 2023 15f009f·Oct 10, 2023 History 6 Commits .cargo chore: cargo compete init Oct 10, 2023 abc322 abc322 a/b/c ...
ABC079_20171118 ABC128_20190526 ABC318 ABC319 ABC320 ABC321 ABC322 ABC323 ABC324 ABC325 ABC326 ABC327 ABC328 ABC329 ABC330 ABC331 ABC332 ABC333 ABC334 ABC335 ABC336 ABC337 ABC338 ABC339 ABC340 ABC341 A.cpp B.cpp C.cpp D.cpp E.cpp ABC342 ABC343 ABC344 ABC345 ABC346 ABC347 ABC...
A - Sequence of Strings Original Link 题目大意: 输入N个字符串,倒序输出。 思想: 签到题。 代码: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 #include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<sstream>#include<vector>#include<queue>#include<stack>...
AtCoder Context ABC 147- C - HonestOrUnkind2 运行要求 运行时间限制:2sec 内存限制:1024MB 原文链接 题目 从1号位到N号位有N个人。他们当中有只说真话的好人,还有说话真假不分的可疑人。i号位的人的发言中进行了Ai个陈述。i号位的人的第j号陈述由Xij和Yij构成。Yij为1的情况的话说明Xij是好人。Yij=0的...
A 水题: #include<iostream> using namespace std; int main(){ string s; cin>>s; if(s=="ABC") cout<<"ARC"<<endl; else cout<<"ABC"<<endl; return 0; } B 水题: 主要是看懂就行,两个单词容易混淆,所以看样例理解比较好 #include<iostream> #define maxn 105 using namespace std; bool...