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

31110011101101011100100100100011010110100010101011 

样例输出

303 
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; }