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開開心心學演算法--Binary String Matching

時間限制:3000 ms  |  記憶體限制:65535 KB 難度: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
輸入
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.
輸出
For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
樣例輸入
3
11
1001110110
101
110010010010001
1010
110100010101011 
樣例輸出
3
0
3 
分析
這道題主要考察字串的配對,只要能正確分析題目意思,按照要求進行輸入輸出,就沒什麼大問題。
我的解:
#include <stdio.h>
#include <string.h>

int main()
{
	int n,count;
	char a[10],b[1000],*p;
	scanf("%d",&n);
	
	while(n--)
	{
		count=0;
		scanf("%s%s",a,b);
		if(strlen(a)>strlen(b) || strlen(a)>10 ||strlen(b)>1000)
			return 0;
		p=b;
		while( (p=strstr(p,a)) !=NULL )
		{
			count++;
			p++;
		}
		printf("%d\n",count);
	}
	return 1;
}

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