NYOJ 5-Binary String Matching

题目链接

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

写这题的时候还不会strstr函数,现在看看strstr好像也能做吧,反正是个很水的题:

 #include<stdio.h>
int main()
{
	int num , i , j , flag ;
	char str[11] , str1[1001];
	scanf("%d" , &num);
	while(num--)
	{
		scanf("%s\n%s" , str , str1);
		for(i = 0 , flag = 0; str1[i] != 0 ; i ++)
		{
			if(str[0] == str1[i])
			{
				for(j = 1 ; str[j] != 0 && str1[i + j] != 0 ; j++)
				{
					if(str[j] != str1[i + j])
						break;
				}
				if(str[j] == 0)
					flag++;
			}
		}
		printf("%d\n" , flag);
	}
	return 0;
}