POJ 1936 All in All

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
*/


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值