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

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 using namespace std; 5 int main(){ 6     int test, j, i; 7     string a, b; 8     int n; 9     cin >> test;10     while(test--){11         n = 0;12         cin >> a >> b;13         int len_a = a.length(), len_b = b.length();14         for(i = 0; i < len_b; i++){15             int k = i;16             for(j = 0; j < len_a; k++,j++){17                 if(b[k] != a[j])18                     break;19             }20             if(j == len_a)21                 n++;22         }23         cout << n << endl;24     }25     return 0;26 }
         
        
       

上面是直接遍历。

以下代码利用find()函数

1 #include 
       
         2 #include 
        
          3 using namespace std; 4  5 int main(){ 6     int t; 7     string a, b; 8     cin >> t; 9 10     while(t--){11         cin >> a >> b;12         int n = 0;13         int index = b.find(a, 0);//返回从0开始找到子串在串中的位置下标14         while(index != b.npos){
     //npos表示不存在15             n++;16             index = b.find(a, index + 1);17         }18         cout << n << endl;19     }20 21     return 0;22 }