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 100005
char s1[MAX],s2[MAX];
int main(){
    while(scanf("%s%s",s1,s2)!=EOF){
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        int i=0;
        int j=0;
        while(1){
            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 0;
}