POJ 1936 All in All(简单的字符串的字符列)

All in All

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21789		Accepted: 8668

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

 解题报告：就是字符串的子序列问题，匹配到最后时只要前面的单词按照顺序依次在后面的字符串中出现即可

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
const int N = 100010;
char a[N], b[N];
int main()
{
    int lena, i, j, lenb;
    while (scanf("%s%s", a, b) != EOF)
    {
        lena = strlen(a);
        lenb = strlen(b);
        for (i = 0, j = 0; i < lenb; ++i)
        {
            if(b[i] == a[j])
            {
                ++j;
            }
        }
        if (j == lena)//匹配到最后时只要前面的单词按照顺序依次在后面的字符串中出现即可
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}