Print all numbers that are divisors of N and are co-prime with the quotient of their division

Given a positive integer N, the task is to print all the numbers, say K, such that K is a divisor of N and K and N / K are coprime.

Examples:

Input: N = 12  
Output: 1 3 4 12  
Explanation:
All numbers K such that it is divisor of N(= 12) and K and N/K are coprime:

1. 1: Since 1 is a divisor of 12 and 1 and 12(= 12/1) are coprime.
2. 3: Since 3 is a divisor of 12 and 3 and 4( = 12/3) are coprime.
3. 4: Since 4 is a divisor of 12 and 4 and 3( = 12/4) are coprime.
4. 12: Since 12 is a divisor of 12 and 12 and 1( = 12/12) are coprime.

Input: N = 100  
Output: 1 4 25 100

Naive Approach: The simplest approach to solve the given problem is to iterate over the range [1, N] and check for each number K whether K is a divisor of N and GCD of K and N/K is 1 or not. If found to be true, then print K. Otherwise, check for the next number.

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the observation that all the divisors are present in pairs. For Example, if N is 100, then all the pairs of divisors are: (1, 100), (2, 50), (4, 25), (5, 20), (10, 10).

Therefore, the idea is to iterate in the range [1, ?N] and check if both the given conditions are satisfied or not i.e., whether any number K is a divisor of N and GCD of K and N/K is 1 or not. If found to be true, then print both the numbers. In the case of two equal divisors, print only one of them.

Below is the implementation of the above approach:  

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to print all numbers
// that are divisors of N and are
// co-prime with the quotient
// of their division
void printUnitaryDivisors(int n)
{
    // Iterate upto square root of N
    for (int i = 1; i <= sqrt(n); i++) {

        if (n % i == 0) {

            // If divisors are equal and gcd is
            // 1, then print only one of them
            if (n / i == i && __gcd(i, n / i) == 1) {
                printf("%d ", i);
            }

            // Otherwise print both
            else {
                if (__gcd(i, n / i) == 1) {
                    printf("%d %d ", i, n / i);
                }
            }
        }
    }
}

// Driver Code
int main()
{
    int N = 12;
    printUnitaryDivisors(N);

    return 0;
}

Python3

# python 3 program for the above approach
from math import sqrt, gcd

# Function to print all numbers
# that are divisors of N and are
# co-prime with the quotient
# of their division
def printUnitaryDivisors(n):

    # Iterate upto square root of N
    for i in range(1,int(sqrt(n)) + 1, 1):
        if (n % i == 0):

            # If divisors are equal and gcd is
            # 1, then print only one of them
            if (n // i == i and gcd(i, n // i) == 1):
                print(i)

            # Otherwise print both
            else:
                if (gcd(i, n // i) == 1):
                    print(i, n // i,end = " ")

# Driver Code
if __name__ == '__main__':
    N = 12
    printUnitaryDivisors(N)

    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

static int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}
// Function to print all numbers
// that are divisors of N and are
// co-prime with the quotient
// of their division
static void printUnitaryDivisors(int n)
{
    // Iterate upto square root of N
    for (int i = 1; i <= (int)Math.Sqrt(n); i++) {

        if (n % i == 0) {

            // If divisors are equal and gcd is
            // 1, then print only one of them
            if (n / i == i && gcd(i, n / i) == 1) {
                Console.Write(i+" ");
            }

            // Otherwise print both
            else {
                if (gcd(i, n / i) == 1) {
                    Console.Write(i + " " +n / i+ " ");
                }
            }
        }
    }
}

// Driver Code
public static void Main()
{
    int N = 12;
    printUnitaryDivisors(N);
}
}

// This code is contributed by SURENDRA_GANGWAR.

Java

// Java program for the above approach
import java.util.*;

class GFG {

    static int gcd(int a, int b)
    {
        return b == 0 ? a : gcd(b, a % b);
    }
    // Function to print all numbers
    // that are divisors of N and are
    // co-prime with the quotient
    // of their division
    static void printUnitaryDivisors(int n)
    {
        // Iterate upto square root of N
        for (int i = 1; i <= (int)Math.sqrt(n); i++) {

            if (n % i == 0) {

                // If divisors are equal and gcd is
                // 1, then print only one of them
                if (n / i == i && gcd(i, n / i) == 1) {
                    System.out.print(i + " ");
                }

                // Otherwise print both
                else {
                    if (gcd(i, n / i) == 1) {
                        System.out.print(i + " " + n / i
                                         + " ");
                    }
                }
            }
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 12;
        printUnitaryDivisors(N);
    }
}

Javascript

<script>

// JavaScript program for the above approach
function gcd( a,  b)
    {
        return b == 0 ? a : gcd(b, a % b);
    }
    // Function to print all numbers
    // that are divisors of N and are
    // co-prime with the quotient
    // of their division
    function printUnitaryDivisors( n)
    {
        // Iterate upto square root of N
        for (var i = 1; i <= Math.sqrt(n); i++) {

            if (n % i == 0) {

                // If divisors are equal and gcd is
                // 1, then print only one of them
                if (n / i == i && gcd(i, n / i) == 1) {
                    document.write(i + " ");
                }

                // Otherwise print both
                else {
                    if (gcd(i, n / i) == 1) {
                        document.write(i + " " + n / i
                                         + " ");
                    }
                }
            }
        }
    }

    // Driver Code
        var N = 12;
        printUnitaryDivisors(N);

// This code is contributed by shivanisingh2110

 </script>

Output:

1 12 3 4

Time Complexity: O(?N*log N)
Auxiliary Space: O(1)