NYOJ - Binary String Matching

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;
    string b;
    int T;

    cin >> T;
    for (int i = 0; i < T; ++i)
    {
        cin >> a >> b;
        int lena = a.size();
        int lenb = b.size();

        int ans = 0;
        for (int i = 0; i <= lenb - lena; ++i)//it is length of string, so watching out the '='
        {
            if (b.substr(i, lena) == a)
                ans++;
        }
        cout << ans << endl;
    }
    return 0;
}

a water question, for training my string STL