UVA - 10340 - All in All （字符串处理！）

Problem E

All in All

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

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 Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output Specification

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

Sample Output

Yes

No

Yes

No

Source: ULM Local Contest

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy :: Non Classical, Usually Harder
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 3. Arrays and Strings :: Exercises
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 1. Algorithm Design :: General Problem Solving Techniques :: Exercises: Beginner
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy - Standard

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms :: Greedy

/*************************************************************************
	> File Name: a.cpp
	> Author: zzuspy
	> Mail: zzuspy@qq.com 
	> Created Time: 2014年12月01日 星期一 20时02分23秒
 ************************************************************************/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#define LL long long
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
using namespace std;

char a[100005], b[100005];      //题目没说多大，我之前只开10005,RE到死啊！！！！

int main()
{
	while(scanf("%s %s", a, b)!=EOF)
	{
		int i, j, lena = strlen(a), lenb = strlen(b);
		if(lena>lenb)
		{
			printf("No\n");
			continue;
		}
		for(i = 0, j = 0; b[j]!='\0'&&i<lena; j++)
		{
			if(lena-i>lenb-j)break;
			if(a[i]==b[j]) i++;
		}
		if(i==lena)printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}