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 }