UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 33318		Accepted: 13901

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

问题链接：UVA10340 POJ1936 ZOJ1970 All in All。入门练习题，用C语言编写程序。

题意简述：输入两个字符串s和t，看s是否是t的子串。t中的字符可以任意删除。

问题分析：顺序匹配字符串即可。

程序说明：（略）

AC的C语言程序如下：

/* UVA10340 POJ1936 ZOJ1970 All in All */

#include <stdio.h>
#include <string.h>

#define MAXN 110000

char s[MAXN], t[MAXN];

int delstrcmp(char *s, char *t)
{
    int i, j, slen, tlen;

    slen = strlen(s);
    tlen = strlen(t);

    for(i=0, j=0; i<slen && j<tlen;) {
        if(s[i] == t[j]) {
            i++;
            j++;
        } else
            j++;
    }

    return i == slen;
}

int main(void)
{
    while(scanf("%s%s", s, t) != EOF)
        printf("%s
", delstrcmp(s, t) ? "Yes" : "No");

    return 0;
}