AtCoderBeginnerContest374(A-E)比赛链接A-Takahashisan2#include<bits/stdc++.h>usingnamespacestd;usingi64=longlong;voidShowball(){strings;cin>>s;intn=s.size();cout<<(s.substr(n-3)=="san"

A-C：惯例是宝宝题，会打暴力就能过哈D：其实也是暴力dfs，有一个double打错成int（我是猪鼻），卡了我很久#include<bits/stdc++.h>usingnamespacestd;constintmaxn=1e3+10,eps=1e-7;intn,s,t;boolvis[10];doublesum=1e8;structNode{ doublex,y,x1,y1;}a[maxn];doub

A.Takahashisan2判断一个字符串是否以san结尾usingnamespacereader;intmain(){strings;cin>>s;if(s[s.length()-1]=='n'ands[s.length()-2]=='a'ands[s.length()-3]=='s'){cout<<"Yes";

ABC374A-Takahashisan2题目传送门代码（签到题）#include<cstdio>#include<cctype>#include<cstring>#include<cmath>#include<queue>usingnamespacestd;intiut(){ intans=0,f=1;charc=getchar(); while(!isdigit(c))f=(c==&

省流版A.判断末三位即可B.逐位判断即可C.枚举所有分组情况即可D.枚举线段顺序、端点顺序即可E.二分答案，发现贵的机器数量不超过\(100\)，枚举求最小花费看是否可行即可F.朴素DP，复杂度分析得到有效时刻不超过\(O(n^2)\)而非\(O(s_i)\)，直接\(DP\)即可A-Takahashi

题目：题解：publicclassSolutionextendsGuessGame{publicintguessNumber(intn){intleft=1,right=n;while(left<right){//循环直至区间左右端点相同intmid=left+(right-left)/2;//防止计算时溢出