AtCoder Beginner Contest 129 ABCD 签到(A、B、C过水已隐藏) View Code E 发现x^y=x+y时,x,y没有同时为1的位。于是数位DP,f[i][0/1]表示到了第i位是否达到上限,发现该位取1有2种方案,取0有1种,大力O(n)DP即可。 View Code F 设计算到x时,答案是ans,于是计算x后,ans=(ans*10i+x)%m,其...
思路:刚开始傻逼傻逼的对每个点往四个方向延伸,然后就获得了一个TLE。从左上对每个点统计从它的左方和上方分别能走多远,从右下对每个点统计从它的右方和下方分别能走多远 然后再对每一个点求一边四个方向的和的最大值 View Code E - Sum Equals Xor 题意:给一个二进制数LL,问有多少对(a,b)(a,b)满...
m;intp[N];intsz[N],e[N];intfind(intx){if(p[x]!=x)p[x]=find(p[x]);returnp[x];}voidsolve(){cin>>n>>m;for(inti=1;i<=n;i++)p[i]=i,sz[i]=1;for(inti=0;i<m;i++){inta,b;cin>>a>>b;intpa=find(a),pb=find(b);if(pa==pb)e[pb]++;else{e[pb]+=e[pa...
Did you enjoy the AtCoder Beginner Contest 128? A Japanese editorial is already out, but unfortunately there is no English editorial, so I translated it into English experimentally. Note that this is an unofficial one; AtCoder has no responsibility for this editorial. Also, I didn't do ...
AtCoder Beginner Contest 400 A-F LHao AtCoder Beginner Contest 401 A-G 简易题解,如果题解中有什么问题可以找我反馈,谢谢!(感觉这次格式搞得有点怪,回头再调下) A.Status Code直接判断即可。 int main(){ int n; cin >> n; if(n >= 200 &… 枫落发表于Atcod... 2.第一行代...
AtCoder Beginner Contest 269「A」「B」「C 二进制枚举」「D 暴力dfs」「E 二分答案」「F 等差数列+推式子」 https://suryxin.blog.csdn.net/article/details/126974824 AtCoder Beginner Contest 269(E-EX) https://www.bilibili.com/opus/707445585544740886 ...
判断图中存在闭环的常用方法——以Atcoder Beginner Contest 285(D - Change Usernames)为例 656 2 24:27 App AtCoder Beginner Contest 370(A ~ F 题讲解) 472 1 06:29 App AtCoder Beginner Contest 354(G 题讲解) 466 1 47:47 App AtCoder Beginner Contest 388(A ~ G 题讲解) 502 0 22:37 ...
AtCoder Beginner Contest 216 个人题解 每篇一图 A题 Signed Difficulty 题目大意: 给出一个小数,根据小数部分改写 \(+,-\) 思路解析: 直接判断即可 AC代码: #include<bits/stdc++.h> using namespace std; int main(){ int x,y; scanf("%d.%d",&x,&y);...
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...
We will hold AtCoder Beginner Contest 199(Sponsored by Panasonic). Start Time: Duration: 100 minutes Number of Tasks: 6 Writer:evima, kyopro_friends,QCFium,satashun,YoshikaMiyafuji Rated range: ~ 1999 If any of the vertices of a component have degree 3 or higher so our total answer is ...