Binary String Matching

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

#include<stdio.h>
#include<string.h>
int main()
{
int n,t;
char a[10],b[1000],*p;
scanf("%d",&n);
while(n--)
{
t=0;
scanf("%s%s",a,b);
p=b;
while( (p=strstr(p,a)) !=NULL )
{
t++;
p++;
}
printf("%d\n",t);
}
return 0;
}

//其它人写的，差不多

#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int main()
{
int n;
char A[11];
char B[1100];
char *p;
cin>>n;
while(n--)
{
int count=0;
cin>>A>>B;
p=strstr(B,A);
while(p)
{
count++;
p=p+1;
p=strstr(p,A);
}
cout<<count<<endl;
}
return 0;
}

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