AtCoderBeginnerContest359A-CountTakahashi有\(n\)个字符串，每个串要么是Takahashi要么是Aoki，问有多少个字符串是Takahashi额....这还要有题解吗(?)#include<iostream>#include<cstring>usingnamespacestd;intmain(){stringa;intn,ans=0;cin>

AtCoderBeginnerContest359这场我赛时打的非常不好，只做出了\(2\)题。A-CountTakahashi签到//Problem:A-CountTakahashi//Contest:AtCoder-UNIQUEVISIONProgrammingContest2024Summer(AtCoderBeginnerContest359)//URL:https://atcoder.jp/conte

A-CountTakahashiQuestion：给你n个单词，要么是Takahashi，要么是Aoki；输出有几个Takahashi即可。Code：#include<bits/stdc++.h>usingnamespacestd;#defineendl'\n'#defineintlonglongtypedeflonglongll;typedefunsignedlonglongull;typedefpair<

https://atcoder.jp/contests/abc359/tasksA-CountTakahashivoidsolve(){ intn; cin>>n; intans=0; while(n--){ strings; cin>>s; if(s=="Takahashi"){ ans++; } } cout<<ans<<endl;}B-

submissionsA,B直接模拟即可。C纵向的距离很好算。有两种情况：横向距离更小。这个直接输出纵向距离。更大。减去纵向的步数。横向距离怎么算？我们考虑把\(s,e\)都移动到方块靠左，然后就是横坐标之和。D简单的dp。设\(dp_{i,msk}\)为到了第\(i\)为，目前前面的状

AtCoderBeginnerContest359(3/6)A-CountTakahashiProblemStatementYouaregivenNNNstrings.Thei

A-CountTakahashi(abc359A)题目大意给定\(n\)个字符串，问有多少个字符串是Takahashi解题思路注意判断比较即可。神奇的代码#include<bits/stdc++.h>usingnamespacestd;usingLL=longlong;intmain(void){ios::sync_with_stdio(false);cin.tie(0);