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 

源代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>char s[1000],t[10];void index(char t[10],char s[1000])//基于数据结构模式匹配BP算法{  int i=0,j=0;  int count=0;  while(i<strlen(s))  {    if(s[i] == t[j])     {      i++;      j++;                   }    else    {      i=i-j+1;      j=0;    }    if(j>=strlen(t))    {            count++;       i=i-j+1;       j=0;         }                  }  printf("%d\n",count);}int main(){  int n;  scanf("%d",&n); // getchar();  while(n--)  {    scanf("%s%s",t,s);      index(t,s);          }  system("pause");  return 0;    }

Binary String Matching