【南理oj】5 - Binary String Matching（水）

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 

来源 网络 上传者 naonao

挺简单的。

代码如下：

#include <cstdio>
#include <cstring>
char base[14];
char pr[1111];
int l1,l2;
bool check(int n)
{
	for (int i=0;i<l1;i++)
	{
		if (base[i]!=pr[n++])
			return false;
	}
	return true;
}
int main()
{
	int u;
	int l;
	int ans;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%s",base);
		scanf ("%s",pr);
		l1=strlen(base);
		l2=strlen(pr);
		l=l2-l1;
		ans=0;
		for (int i=0;i<=l;i++)
		{
			if (check(i))
				ans++;
		}
		printf ("%d\n",ans);
	}
	return 0;
}