All in All UVA
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https://vjudge.net/problem/UVA-10340
对s 和 t 扫一遍
如果 s[i]==t[j] 则i++ ,j++ 代表这个字符得到 之后寻找下面的字符
否则就j++
这样 如果 i走到了字符串末尾
那么 证明s可以从t得到
#include <iostream>#include <cstring>#include <iomanip>#include <cstdio>#include <string>#include <algorithm>#include <queue>#include <cmath>#include <map>using namespace std;int main(){ string s,t; while(cin>>s>>t) { int j=0,i=0,sum=0; while(t[i]&&s[j]) { if(s[j]==t[i]) { i++; j++; } else i++; } if(j==strlen(&s[0])) cout<<"Yes"<<endl; else cout<<"No"<<endl; }}
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