算法竞赛入门经典 第三章 uVa10340 - All in All

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

解题难点在于字符串长度，最开始只用了1000一直通不过

#include
#include
using namespace std;
char str1[1000000];
char str2[1000000];
int main()
{	
	while (scanf("%s%s", str1, str2) != EOF)
	{
		int sign = 0;
		int count = 0;
		for (int i = 0; i < strlen(str1); i++)
		{
			int j;
			for (j = sign; j < strlen(str2); j++)
			{
				if (str1[i] == str2[j])
				{
					sign = j+1;
					count++;
					break;
				}
			}
			if (j == strlen(str2))
				break;
		}
		if (count==strlen(str1))
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}