NYOJ5——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 

#include<iostream>//我用模式匹配算子串出现的次数
#include<string>
using namespace std;

int Match(string pat,string sat)
{
    int count=0;
    int i=0; 
    int m=pat.length(),n=sat.length();
      while(i<=(n-m))
      {
           int j=0;
           while((sat[i]==pat[j])&&(j<pat.length()))
           {
              i++;
              j++;

           }
           if(j==pat.length())
           {
              count++;
           }
           i=i-j+1;
      }
      
  
   return count;
}
int main()
{
  
   int num;   
   cin>>num;
   string  pat,sat;
   while(num--)
   {
      int p=0,s=0;
      int count;
      cin>>pat;
      cin>>sat;

      count=Match(pat,sat);
      cout<<count<<endl;
   }
      
   return 0;
}