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
代码实现：
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
 int n;
 char a[11],b[1002];
 cin>>n;
 while(n--)
 {
  int sum=0,temp,i,j;
  cin>>a>>b;
  for(j=0;j<strlen(b);j++)
  {
   temp=1;
   //从每个字母开始比较
   for(i=0;i<strlen(a);i++)
   {
    if(b[i+j]!=a[i])
     temp=0;
   }
   sum+=temp; 
  }
  cout<<sum<<endl;
 }
 return 0;
 }