NYOJ5 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 <cstring>
using namespace std;

int main()
{
    int T;
    char a[];
    char b[];
    int i;
    int j;
    int count;
    int temp;
    int a_length;
    int b_length;

    cin >> T;
    while (T--)
    {
        cin >> a >> b;

        count = ;
        a_length = strlen(a);
        b_length = strlen(b);
        for (i = ;i < b_length-a_length+;i++)
        {
            temp = i;
            for (j = ;j < a_length;temp++,j++)
            {
                if (b[temp] != a[j])
                {
                    break;
                }
            }

            if (j >= a_length)
            {
                count++;
            }
        }
        cout << count << endl;
    }
    return ;
}