nyoj 5 Binary String Matching（string)

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>
 #include <string>
 #include <algorithm>
 using namespace std;
 int main(){
     int test, j, i;
     string a, b;
     int n;
     cin >> test;
     while(test--){
         n = ;
         cin >> a >> b;
         int len_a = a.length(), len_b = b.length();
         for(i = ; i < len_b; i++){
             int k = i;
             for(j = ; j < len_a; k++,j++){
                 if(b[k] != a[j])
                     break;
             }
             if(j == len_a)
                 n++;
         }
         cout << n << endl;
     }
     return ;
 }

上面是直接遍历。

以下代码利用find()函数

 #include <iostream>
 #include <string>
 using namespace std;

 int main(){
     int t;
     string a, b;
     cin >> t;

     while(t--){
         cin >> a >> b;
         int n = ;
         int index = b.find(a, );//返回从0开始找到子串在串中的位置下标
         while(index != b.npos){//npos表示不存在
             n++;
             index = b.find(a, index + );
         }
         cout << n << endl;
     }

     return ;
 }