NYOJ 5 Binary String Matching

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 

 
#include<iostream>
#include<string>
using namespace std;
int main()
{
    string a,b;
    int n;cin>>n;
    while(n--)
    {
        cin>>a>>b;
        int count=0;
        unsigned int num=b.find(a,0);
        while(num!=string::npos)
        {
            count++;
            num=b.find(a,num+1);
        }
        cout<<count<<endl;
    }
}