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

31110011101101011100100100100011010110100010101011 

样例输出

303 

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int times=scanner.nextInt();String cc=scanner.nextLine();while(times--!=0){String str1=scanner.nextLine();String str2=scanner.nextLine();int count=0;for(int i=0;i<str2.length()-str1.length()+1;i++){String temp=str2.substring(i,i+str1.length());if(temp.equals(str1))count++;}System.out.println(count);}}}