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

題目大意：輸入兩個字串A，B，只包含0或者1，有多少次A出現在B的字串中，例如B是1001110110，A是11，有3次A出現在B中

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 

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();//n=3
		
		String str;
		String pattern;
		//迴圈3次
		while(n > 0){
			pattern = sc.next();//11
			str = sc.next();//1001110110
			
			int index = 0;//記錄pattern在str中出現的索引
			int count = 0;//記錄出現的次數，預設=0
			
			while(index != -1){
				index = str.indexOf(pattern);//index = 3
				if(index != -1){
					//說明存在pattern
					count ++ ;//count = 1
					str = str.substring(index+1);//此時str = 110110
				}
			}
			
			System.out.println(count);
			
			
			n--;
		}
	}
}