Count all pairs of divisors of a number N whose sum is coprime with N

Given an integer N, the task is to count all pairs of divisors of N such that the sum of each pair is coprime with N.

Examples:

Input: N = 24
Output: 2
Explaination:
There are 2 pairs (1, 24) and (2, 3) whose sum is coprime with 24

Input: 105
Output: 4
Explaination:
There are 4 pairs (1, 105), (3, 35), (5, 21) and (7, 15) whose sum is coprime with 105

Approach:



To solve the problem mentioned above we can easily calculate the result by finding all divisors in √N complexity, and check for each pair, whether its sum is coprime with N or not.

Below is the implementation of the above approach:

C++

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// C++ program to count all pairs
// of divisors such that their sum
// is coprime with N
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate GCD
int gcd(int a, int b)
{
    if (b == 0)
        return a;
  
    return (gcd(b, a % b));
}
  
// Function to count all valid pairs
int CountPairs(int n)
{
  
    // Initialize count
    int cnt = 0;
  
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            int div1 = i;
  
            int div2 = n / i;
  
            int sum = div1 + div2;
  
            // Check if sum of pair
            // and n are coprime
            if (gcd(sum, n) == 1)
  
                cnt += 1;
        }
    }
  
    // Return the result
    return cnt;
}
  
// Driver code
int main()
{
  
    int n = 24;
  
    cout << CountPairs(n) << endl;
  
    return 0;
}

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Java

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// Java program to count all pairs
// of divisors such that their sum
// is coprime with N
import java.util.*;
  
class GFG{
  
// Function to calculate GCD
public static int gcd(int a, int b)
{
    if (b == 0)
        return a;
  
    return (gcd(b, a % b));
}
  
// Function to count all valid pairs
public static int CountPairs(int n)
{
  
    // Initialize count
    int cnt = 0;
  
    for(int i = 1; i * i <= n; i++) 
    {
       if (n % i == 0)
       {
           int div1 = i;
           int div2 = n / i;
           int sum = div1 + div2;
             
           // Check if sum of pair
           // and n are coprime
           if (gcd(sum, n) == 1)
               cnt += 1;
       }
    }
  
    // Return the result
    return cnt;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 24;
  
    System.out.println(CountPairs(n));
}
}
  
// This code is contributed by equbalzeeshan

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Python3

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# Python3 program to count all pairs
# of divisors such that their sum
# is coprime with N
import math as m
  
  
# Function to count all valid pairs
def CountPairs(n):
      
    # initialize count 
    cnt = 0
      
    i = 1
      
    while i * i <= n :
          
        if(n % i == 0):
              
            div1 = i
            div2 = n//i
              
            sum = div1 + div2;
              
            # Check if sum of pair
            # and n are coprime
            if( m.gcd(sum, n) == 1):
                cnt += 1
          
        i += 1
      
    # Return the result
    return cnt
      
  
# Driver code 
n = 24
  
print(CountPairs(n))

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C#

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// C# program to count all pairs of
// divisors such that their sum
// is coprime with N
using System;
  
class GFG {
  
    // Function to find gcd of a and b
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
  
    // Function to count all valid pairs
    static int CountPairs(int n)
    {
  
        // Initialize count
        int cnt = 0;
  
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                int div1 = i;
  
                int div2 = n / i;
  
                int sum = div1 + div2;
  
                // Check if sum of pair
                // and n are coprime
                if (gcd(sum, n) == 1)
                    cnt += 1;
            }
        }
        // Return the result
        return cnt;
    }
  
    // Driver method
    public static void Main()
    {
        int n = 24;
  
        Console.WriteLine(CountPairs(n));
    }
}

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Output:

2

Time Complexity: O(√N * log(N))

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Improved By : equbalzeeshan