NUC1657 All in All【字符串匹配】

All in All

时间限制: 1000ms 内存限制: 30000KB

通过次数: 1总提交次数: 1

问题描述

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.

输入描述

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.

输出描述

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

样例输入

sequence subsequenceperson compressionVERDI vivaVittorioEmanueleReDiItaliacaseDoesMatter CaseDoesMatter

样例输出

YesNoYesNo

来源

Ulm Local 2002

问题分析：（略）

这个问题和《UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】》是同一个问题，代码拿过来用就AC了。

程序说明：参见参考链接。

参考链接：UVA10340 POJ1936 ZOJ1970 All in All【字符串匹配】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* UVA10340 POJ1936 ZOJ1970 All in All */#include <stdio.h>#include <string.h>#define MAXN 110000char 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\n", delstrcmp(s, t) ? "Yes" : "No");    return 0;}