POJ 1936 All in All ---- 串匹配

All in All

Time Limit: 1000MS
Memory Limit: 30000KTotal Submissions: 28812
Accepted: 11931

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 subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatter

Sample Output

YesNoYesNo

Source

Ulm Local 2002

简单模式匹配：

#include <stdio.h>#include <string.h>char s1[100010],s2[100010];int main(){    while(~scanf("%s%s",s1,s2))    {        char *p = s1,*q = s2;        while(*p != '\0' && *q != '\0')        {            if(*p == *q)            {                p++;            }            q++;        }        if(*p == '\0')            printf("Yes\n");        else            printf("No\n");    }    return 0;}