题目链接E.连通块的搜索debug：用不上回溯，把连通块的贡献全部加起来#include<bits/stdc++.h>usingnamespacestd;intn,m;intg[1010][1010];boolvis[1010][1010];intma,tmp;intdx[5]={-1,1,0,0};intdy[5]={0,0,1,-1};voiddfs(intx,inty){tmp+=g[x][y];

题目链接A.直接比较输入字符串和已知字符串有几个不同即可#include<bits/stdc++.h>usingnamespacestd;voidsolve(){strings;cin>>s;intans=0;stringt="codeforces";for(inti=0;i<10;i++){if(s[i]!=t[i])ans++;}cout<&

很久没写题目了，划点水题A.LoveStory#include<bits/stdc++.h>usingnamespacestd;typedeflonglongLL;typedefpair<LL,LL>PII;constLLMAXN=1e18;constLLN=1e6,M=4002;constLLmod=1e9+7;intmain(){//cin.tie(0);cout.tie(0);ios

A.LoveStory#include<bits/stdc++.h>usingnamespacestd;#defineintlonglongintread(){intx=0,f=1,ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if

A.LoveStory题意:给定n个长度为10的字符串，问其与codeforces字符串的对应下标字母不同的个数。分析:对于每个字符串从前往后依次和“codeforces”对应字符比较然后统计不同字母数即可code:#include<bits/stdc++.h>usingnamespacestd;intmain(){ std::ios::sync_wit

CodeforcesRound871(Div.4)A-LoveStory#include<bits/stdc++.h>usingnamespacestd;typedefpair<int,int>PII;typedefpair<string,int>PSI;constintN=1e5+5,INF=0x3f3f3f3f,Mod=1e6;constdoubleeps=1e-6;typedeflonglongll;i

Codeforces871div4ABC简单题D题意每次操作可以将当前的数分成两份，一份是\(\frac{1}{3}\)，一份是\(\frac{2}{3}\)，问当前数n可否进行若干次操作，最终出现一份大小为m的片。递归一下就好了，数据最大才\(10^7\)代码voiddfs(intx){ if(x==m){flag=1;return;} if(flag)re