[POJ1936]All in All

[POJ1936]All in All

试题描述

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

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

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

输入示例

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

输出示例

Yes
No
Yes
No

数据规模及约定

见“输入”

题解

求一个串是不是另一个串的子序列，贪心匹配就行了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

#define maxn 100010
int n, m;
char S1[maxn], S2[maxn];

int main() {
	while(scanf("%s%s", S1 + 1, S2 + 1) != EOF) {
		n = strlen(S1 + 1); m = strlen(S2 + 1);
		if(n > m){ puts("No"); continue; }
		int i = 1, j = 1;
		for(; i <= n; i++) {
			for(; S1[i] != S2[j] && j <= m; j++) ;
			if(j > m) break;
			j++;
		}
		if(i > n) puts("Yes");
		else puts("No");
	}

	return 0;
}