UVA - 10340 - All in All
题目:https://cn.vjudge.net/problem/UVA-10340
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; char s[1000005],t[1000005]; int main() { int k,n; while(scanf("%s%s",s,t)!=EOF) { n=k=0; for(int j=0;s[j]!='\0';j++) { for(int i=k;t[i]!='\0';i++) { if(s[j]==t[i]) { // cout<<j<<" "<<i<<endl; // cout<<s[j]<<" "<<t[i]<<endl; n++; k=i+1; break; } } } if(n==strlen(s)) cout<<"Yes"<<endl; else cout<<"No"<<endl; } return 0; }
思路:
两个字符串要顺序查找,找到对应的一个字符后,两者只能继续往后找,注意循环的位置