POJ 1936 All in All（模拟）

All in All

题目链接：http://poj.org/problem?id=1936

题目大意：判断从字符串s2中能否找到子串s1。字符串长度为10W。

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

分析：这明明是模拟题，有人竟然把它归为动态规划，是要用LCS做吗

代码如下：

 # include<stdio.h>
 # include<string.h>
 # define MAX
 char s1[MAX],s2[MAX];
 int main(){
     while(scanf("%s%s",s1,s2)!=EOF){
         int len1 = strlen(s1);
         int len2 = strlen(s2);
         int i=;
         int j=;
         while(){
             if(i==len1){
                 printf("Yes\n");
                 break;
             }
             else if(j==len2){
                 printf("No\n");
                 break;
             }
             if(s1[i]==s2[j])
                 i++,j++;
             else
                 j++;
         }
     }
     return ;
 }