題目如下：

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

輸入

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.

輸出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

樣例輸入

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出

3
0
3 

演算法分析：

利用Java自帶的substring(beginningindex，endindex)提取字串B中的字串，利用match函式比較，若相同則計數加1.

程式碼如下：

package binarymatch;
import java.util.*;

public class Main {
	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);
		int num = 0;
	//	System.out.println("輸入num");
		num = input.nextInt();
		String[] A = new String[num];
		String[] B = new String[num];
		for(int index = 0; index < num ; index++)
		{
//		System.out.println("輸入A");
		A[index] = input.next();
//		System.out.println("輸入B");
		B[index] = input.next();
		}
		for(int index = 0; index < num ; index++)
		{
			System.out.println(binary_string_match(A[index],B[index]));
		}
		
	}
	public static int binary_string_match(String A, String B)
	{
		int ret = 0;
		int  lengtha = A.length();
		int  lengthb = B.length();
		for(int index = 0; index < (lengthb-lengtha+1); index++)
		{
			String sub = new String();
			sub = B.substring(index, index+lengtha);
			if(sub.matches(A))
				ret++;
			
		}
		
		return ret;
		
	}

}