1.链接：http://poj.grids.cn/practice/1936/

2.题目：

总时候限制:

1000ms

内存限制:

65536kB

描述

You have devised a new encryption technique which encodes a message by ing 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 ed 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 t such
that the concatenation of the remaining characters is s.

输入

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.

输出

For each test case output Yes， if s is a subsequence of t，otherwise output No.

样例输入

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

样例输出

Yes

No

Yes

No

3.代码：

 1 ＃include <iostream>

 2 ＃include <cstdio>

 3 ＃include <cstring>

 4 ＃include <cstdlib>

 5 using namespace std;

 6 int main（）

 7 {

 8     //freopen（F:\\input.txt，r，stdin）;

 9     

10     

11     char s[100001]，t[100001];

12     while（scanf（％s ％s，s，t） != EOF）

13     {

14         int len1 = strlen（s）;

15         int len2 = strlen（t）;

16         int i，j;

17         

18         i = 0;j = 0;

19         while（i < len1 && j < len2）

20         {

21             if（s[i] == t[j]） i++;

22             j++;

23         }

24         if（i >= len1） cout<<Yes<<endl;

25         else cout<<No<<endl;

26     }

27     return 0;

28 }

4.思路：

1.应用scanf（％s ％s，s，t） != EOF断定是否停止