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