OpenJudge/Poj 1936 All in All

1.链接地址：
http://poj.org/problem?id=1936 http://bailian.openjudge.cn/practice/1936
2.题目：

All in All

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 26651		Accepted: 10862

Description
You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a
subsequence of t, i.e. if you can remove characters from t such that the
concatenation of the remaining characters is s.

Input
The
input contains several testcases. Each is specified by two strings s, t
of alphanumeric ASCII characters separated by whitespace.The length of s
and t will no more than 100000.
Output
For each test case output "Yes", if s is a subsequence of t,otherwise output "No".
Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

Source
Ulm Local 2002

3.思路：

 

4.代码：

1 #include "stdio.h"
2 //#include "stdlib.h"
3 #include "string.h"
4 #define NUM 100002
5 char s[NUM],t[NUM];
6 int main()
7 {
8     int i,j;
9     int len_s,len_t;
10     while(scanf("%s%s",s,t) != EOF)
11     {
12         len_s=strlen(s);
13         len_t=strlen(t);
14         i=0;j=0;
15         while(i<len_s && j<len_t)
16         {
17            if(s[i]==t[j]) i++;
18            j++;
19         }
20         //printf("%s%s\n",s,t);
21         if(i>=len_s) printf("Yes\n");
22         else printf("No\n");
23     }
24     //system("pause");
25     return 0;
26 }