POJ 1963：All in All

POJ 1963：All in All

All in All Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 27707 Accepted: 11381

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

簡單的字符串問題。。判斷後一個字符串是否包含前一個。。好久沒0MS 1A 了 。。。OTL。。。

#include
#include
#include
#include
#include
#include
using namespace std;

const int maxn = 100000 + 50;
char a[maxn];
char b[maxn];

int main()
{
    while(scanf("%s%s", a, b)==2)
    {
        int x = strlen(a);
        int y = strlen(b);
        int t = 0;
        int judge = 0;
        if( y