Count ordered pairs with product less than N in C++

We are given a number N. The goal is to find ordered pairs of positive numbers such that their product is less than N.

We will do this by starting from i=1 to i<N and j=1 to (i*j)<N. Then increment count.

Let’s understand with examples.

Input 

N=4

Output 

Ordered pairs such that product is less than N:5

Explanation 

Pairs will be (1,1) (1,2) (1,3) (2,1) (3,1)

Input 

N=100

Output 

Ordered pairs such that product is less than N: 473

Explanation 

Pairs will be (1,1) (1,2) (1,3)....(97,1), (98,1), (99,1). Total 473.

Approach used in the below program is as follows

• We take integer N.

• Function productN(int n) takes n and returns the count of ordered pairs with product < n

• Take the initial variable count as 0 for pairs.

• Traverse using two for loops for making pairs.

• Start from i=1 to i<n. And j=1 to (i* j)<n.

• Increment count by 1.

• At the end of all loops count will have a total number of such pairs.

• Return the count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int productN(int n){
   int count = 0;
   for (int i = 1; i < n; i++){
      for(int j = 1; (i*j) < n; j++)
         { count++; }
   }
   return count;
}
int main(){
   int N = 6;
   cout <<"Ordered pairs such that product is less than N:"<<productN(N);
   return 0;
}

Output

If we run the above code it will generate the following output −

Ordered pairs such that product is less than N:10