ACM刷题: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

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

来源：网络

上传者：naonao

代码

#include <stdio.h>
int  main(){
  int n,i,j=0,k=0,l=0;
  scanf("%d",&n);
  getchar();
  char a[10];
  char b[1000];
   for(i=0;i<n;i++){
    scanf("%s",a);
    getchar();
    scanf("%s",b);
    getchar();
    while(b[j]!='\0'){
      while(a[k]!='\0'){
        if(a[k]!=b[j+k]){
        break;
        }
        k++;
        if(a[k]=='\0'){
          l++;
          break;
        }
      }
      j++;
      k=0;
    }
    printf("%d\n",l);
    k=0;
    j=0;
    l=0;
   }
}