pku_1936 All in All

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#concatenation #encryption #string #output #each #input

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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。通过遍历两个字符串并比较字符来实现这一功能。

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题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=1936

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
================================================================

题目类型：简单题
算法：只要在第二个字符串中按照顺序找到第一个字符串中所有字符即可
#include<iostream>
#include<string>
using namespace std;

bool judge(const string &s1,const string &s2)
{
	int la=s1.length();
	int lb=s2.length();
	int i=0,j=0;
	while(i<la&&j<=lb)
	{
		if(s1[i]==s2[j]) 
		{
			++i;
			++j;
		}
		else ++j;
	}
	if(i<la) return false;
	else return true;
}

int main()
{
	string s1,s2;
	while(cin>>s1>>s2)
	{
		bool flag=judge(s1,s2);
		if(flag) cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}
	return 0;
}