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

代码：

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
    int n,ans,num;
    string s,p;
    cin>>n;
    while(n--)
    {
        cin>>p>>s;
        ans=0,num=0;
        ans=s.find(p,0);
        while(ans!=string::npos)
        {
            ans=s.find(p,ans+1);
            num++;
        }
        cout<<num<<endl;
    }
    return 0;
}