算法竞赛入门经典第三章 uVa10340 - All in All_you have devised a new encryption technique which -优快云博客

本文链接：https://blog.youkuaiyun.com/ColorlessSilver/article/details/42045463

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。通过逐字符比对的方法，该算法能在较短时间内得出结论，适用于多种应用场景。

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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

解题难点在于字符串长度，最开始只用了1000一直通不过

#include<cstdio>
#include<cstring>
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;
}