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

代码实现：

#include 
     
      #include 
      
       int main(){    int n,i,j;    int temp;    /** 保存A,B字符串 **/    char A[11];    char B[1001];    //保存个数    int count = 0;    scanf("%d",&n);    while(n--){        getchar();        count = 0;        scanf("%s",A);        scanf("%s",B);        for(i = 0;B[i] != '\0';i++){            if(B[i] == A[0]){                //printf("0 == yes\n");                int temp = i;                for(j = 0;A[j] != '\0' && B[temp] != '\0';j++,temp++){                    if(B[temp] != A[j]){                        //printf("break\n");                        break;                    }                    //printf("==\n");                }                if(A[j] == '\0')                    count++;            }        }        printf("%d\n",count);    }    return 0;}