南阳oj 一种排序 题目5



/*
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*/
#include<stdio.h>
#include<stdlib.h>
#define N 1000
#include<string.h>
using namespace std;
char a[N],b[10];
int main()
{
 int n;
 scanf("%d",&n);
 while(n--)
 {
  int k1,k2,sum=0,l,i,j;
  scanf("%s",b);
  scanf("%s",a);
  k1=strlen(a);
  k2=strlen(b);
  for(i=0;(i+k2-1)<k1;i++)
  {
    l=i;//记录i的值
    for(j=0;j<k2;j++)
    {
           if(b[j]==a[i]) i++;//如果相等i也要往后走
           else break;//不想等就结束
    }
          i=l;
          if(j==k2) sum++;//j之和k2相等就是说明全部匹配
  }
  printf("%d\n",sum);
 }
 return 0;
}