AtCoder Beginner Contest 292 题解(A-Ex) A - CAPS LOCK (Diff. 7) 简要题意: Sample: Solution: Code: B - Yellow and Red Card (Diff. 39) 简要题意: Sample: Solution: Code: C - Four Variable (Diff.444) 简要题意: Sample: Solution: Code: D - Unicyclic Components(Diff.579) 简要题意...
# AtCoder Beginner Contest 292(E,F,G) ## E(图) [E](https://atcoder.jp/contests/abc292/tasks/abc292_e) 这个题的大意是给你一个有$n$个点,$m$条单向边的图,然后我们需要把满足下面条件的点连接在一起,
AtCoder Beginner Contest 402(A-F详细题解) A 思路:我们直接输出字符串中的大写字母即可。 代码:#include<bits/stdc++.h> using namespace std; #define int long long #define N 500010 signed main(){ string s;cin>>… 秋日薄雾 基于Encoder-Decoder框架实现Seq2Seq模型 张浩 Encoder2Pre...
试图引流.jpgF把题看错了,搞了半天。, 视频播放量 488、弹幕量 0、点赞数 20、投硬币枚数 1、收藏人数 4、转发人数 0, 视频作者 grass8cow, 作者简介 菜菜,相关视频:AtCoder Beginner Contest 293 7题,rk89,真耻辱。,AtCoder Beginner Contest 349 rk11,AtCoder Begi
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 406 简易题解,如果题解中...
AtCoder Beginner Contest 287 A-D https://zhuanlan.zhihu.com/p/601517629 AtCoder Beginner Contest 287(A~E) https://blog.csdn.net/weixin_62802134/article/details/128794767 AtCoder Beginner Contest 287 A-F讲解 https://www.bilibili.com/video/BV12d4y1J7JH/ ...
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+1为aia_iai的中位数,bk+1...
We will hold AtCoder Beginner Contest 199(Sponsored by Panasonic). ,satashunYoshikaMiyafuji Rated range: ~ 1999 The point values will be 100-200-300-400-500-600. We are looking forward to your participation! If any of the vertices of a component have degree 3 or higher so our total answ...
AISing Programming Contest 2021(AtCoder Beginner Contest 202) 简单题解记录,补题链接:HereA-ThreeDice水题,问给定三次摇色子的正面,请问3次结果以后相对面的点数和cout<<(21-a-b-c)<<"\n";B-180°水题,首先反转字符串,然后遇到6或者9则相互替代即可voidsolve(){
题意: 题解 利用一个数能被3整除当且仅当其各位之和sum能被3整除。 如果sum本身能被3整除,则不需要删除。 否则统计原数的每一位数%3后的个数,比较%3 =1与%3 =2 的个数,有两种方法可以使其sum变为 %3 =0: %3=1 与%3=2,相互抵消,还剩下的差值即为答案。