Binary String Matching（str.find（））

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
这个用C++里面的str.find()函数就可以实现！
注意一下结束的标志string::npos！具体我也不知道！

#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
	int n,flag,b;
	string s1,s2;
	scanf("%d",&n);
	while (n--)
	{
		flag=0;
		cin>>s1>>s2;
		b=s2.find(s1,0);
		while (b!=string::npos)
		{
			flag++;
			b=s2.find(s1,b+1);
		}
		printf("%d\n",flag);
	}
}