POJ 1936 All in All

最新推荐文章于 2013-09-08 23:13:29 发布

原创最新推荐文章于 2013-09-08 23:13:29 发布 · 492 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#concatenation #encryption #output #each #input

ACM 同时被 2 个专栏收录

41 篇文章

订阅专栏

搜索

1 篇文章

订阅专栏

本文介绍了一个简单的算法，用于判断一个字符串是否为另一个字符串的子序列。通过贪心匹配的方法，该算法有效地验证了原始消息是否被正确编码在最终字符串中。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

All in All

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 21411 Accepted: 8511

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

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

#define MAXN 100001
char s[MAXN], t[MAXN];
int main(){
	int i, j, k;
	while(scanf("%s %s", s, t) != EOF){
		for (i = j = 0; s[i] && t[j]; i++, j++){
			while(t[j] && s[i] != t[j]) j++;
			if (!t[j]) break;
		}
		if (!s[i]) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}
/*
贪心地匹配，让前面的尽量早匹配到，好留给更多的机会给后面
*/