UVa-10340

You have devised a new encryption technique which encodes a message by inserting between its charactersrandomly generated strings in a clever way. Because of pending patent issues we will not discuss indetail 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 finalstring.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can removecharacters 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 ASCIIcharacters separated by whitespace. Input is terminated by EOF.

Output

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

思路：只要两字符串有相应的两字母匹配了，两字符串的索引就都加1

欢迎交流：

解法一：

#include<stdio.h>
#include<string.h>

char s[100005], t[100005];

int main()
{
	while (scanf("%s%s", s, t) !=EOF)
	{
		int lens = strlen(s);
		int lent = strlen(t);
		int count = 0;
		int j = 0;
		for (int i = 0; i < lens; i++)
		{
			for (; j < lent; j++)//不要在此处定义j，否则break后j又恢复为0
			{
				if (s[i] == t[j])
				{
					count++;
					j++;//一会break;后不会执行本for循环的j++
					break;
				}
			}
		}
		if (count == lens)printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

解法二：

参考：http://blog.youkuaiyun.com/mobius_strip/article/details/8225616

#include <iostream>  
#include <cstdlib>  
#include <cstring>  
  
using namespace std;  
  
char str[ 100005 ];  
char val[ 100005 ];  
  
int main()  
{  
    while ( cin >> val >> str ) {  
        int l1 = strlen(val);  
        int l2 = strlen(str);  
          
        if ( l1 > l2 ) {  
            cout << "No" << endl;  
            continue;  
        }  
          
        int s = 0;  
        for ( int i = 0 ; i < l2 ; ++ i )   
            if ( val[s] == str[i] && s ++ == l1-1 )// val[s] == str[i]成立则执行s ++ == l1-1，否则不执行 
                break;  
          
        if ( s == l1 )  
            cout << "Yes" << endl;  
        else cout << "No" << endl;  
    }  
    return 0;  
}