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=;
         unsigned int num=b.find(a,);
         while(num!=string::npos)
         {
             count++;
             num=b.find(a,num+);
         }
         cout<<count<<endl;
     }
 }