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
1 import java.util.Scanner; 2 3 public class Main { 4 public static void main(String[] args) { 5 Scanner scanner=new Scanner(System.in); 6 int T; 7 String A; 8 String B; 9 int startPoint; 10 int count; 11 12 T=scanner.nextInt(); 13 while(T!=0){ 14 T--; 15 startPoint=0; 16 count=0; 17 18 A=scanner.next(); 19 B=scanner.next(); 20 21 while(true){ 22 startPoint=B.indexOf(A,startPoint); 23 24 if(startPoint==-1) 25 break; 26 27 count++; 28 startPoint++; 29 } 30 31 System.out.println(count); 32 } 33 } 34 }