POJ1936-All in All(java版)

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#java #POJ #字符串

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本文探讨了一种新的加密技术，通过插入随机生成的字符串来编码消息。文章提供了一个程序来验证消息是否确实被编码在最终字符串中，通过判断一个字符串是否为另一个字符串的子序列。实现了一个算法来高效地解决此问题，并提供了详细步骤和示例。

All in All

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 27315		Accepted: 11169

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

题目大意是判断s是否是t的子串。

按照题意模拟就行！~

import java.util.Scanner;

public class P1936_AllInAll {
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while(in.hasNext()){
			String s=in.next();
			String t=in.next();
			int slen=s.length();
			int tlen=t.length();
			int j=-1;
			int i=0;
			boolean flag=false;
			if(slen==1){
				System.out.println(t.contains(s)?"Yes":"No");
			}
			else{
				for(;i<slen&&j<tlen;i++){
					while(++j<tlen&&t.charAt(j)!=s.charAt(i));
				}
				System.out.println(i==slen&&j!=tlen?"Yes":"No");
			}
		}
	}
}