[题解]AtCoder Beginner Contest 392(ABC392) A~G A - Shuffled Equation显然只有最大值可能被相乘得到,所以对aa从小到大排序,判断a[0]×a[1]=a[2]a[0]×a[1]=a[2]是否成立即可。时间复杂度O(1)O(1)。点击查看代码 #include<bits/stdc++.h>...
给定M个数,每个数都在1到N中,且两两不同。 列出所有1到N中没有的数字。 用一个数组标记那些数字出现过即可。 #include<iostream>usingnamespacestd;constintMAX=1010;boolvis[MAX];intn,m;intmain(){ cin>>n>>m; cout<<n-m<<endl;for(inti=1;i<=m;++i) {inta; cin>>a; vis[a]=true; }f...
AtCoder Beginner Contest 403 A-G 简易题解,如果题解中有什么问题可以找我反馈,谢谢! A.Odd Position Sum直接 for 循环输入,只对奇数求和即可。int main(){ int n; cin >> n; int res = 0; for(int i = 1; i… 枫落发表于Atcod... AtCoder Beginner Contest 401 A-G 简易题解,如果...
We will hold Japan Registry Services (JPRS) Programming Contest 2025#1 (AtCoder Beginner Contest 392). Contest URL:https://atcoder.jp/contests/abc392 Start Time:http://www.timeanddate.com/worldclock/fixedtime.html?iso=20250208T2100&p1=248 ...
AtCoder Beginner Contest 392 bigseal acm废物 1 人赞同了该文章A Shuffled Equation 排个序判断一下void solve(){ int n = 3; std::vector<int> a(n); cin >> a[0] >> a[1] >> a[2]; sort(a.begin(), a.end()); if(a[0] * a[1] == a[2]) cout << "Yes" << '\n'; ...
We will hold AtCoder Beginner Contest 396. Contest URL: https://atcoder.jp/contests/abc396 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20250308T2100&p1=248 Duration: 100 minutes Writer: toam, sounansya Tester: physics0523, yuto1115 Rated range: ~ 1999 The point...
判断图中存在闭环的常用方法——以Atcoder Beginner Contest 285(D - Change Usernames)为例 737 0 30:29 App AtCoder Beginner Contest 392(A ~ G 题讲解) 559 0 21:08 App Atcoder Beginner Contest 338 289 0 17:35 App Atcoder Beginner Contest 348(A ~ F 讲解,无 D) 1326 0 22:18 App ...
AtCoder Beginner Contest 283 E - Don't Isolate Elements https://www.bilibili.com/video/BV1oD4y1L76r/ https://www.cnblogs.com/zltzlt-blog/p/17344192.html AtCoder Beginner Contest 283 E. Don‘t Isolate Elements(预支下一行状态的状压dp) ...
AtCoder Beginner Contest 169(题解) AtCoder Beginner Contest 169(题解)E - Count Median结论题给定nnn个xi∈[ai,bi]x_i\in[a_i,b_i]xi∈[ai,bi],求中位数的个数。定义:k=⌊n2⌋k=\lfloor\dfrac{n}{2}\rfloork=⌊2n⌋,对a,ba,ba,b进行排序后,为ak+1a_{k+1}ak...
AtCoder Beginner Contest 193 部分题解 E - Oversleeping 求是否存在\(t\)满足\(t=t_1(mod (2X+2Y)) and t=t_2(mod (P+Q))\) 注意到\(Q\)和\(Y\)非常小,直接枚举套个\(exCRT\)就行了(虽然赛场上没看出来,\(exCRT\)也忘了记得快速乘...