Generate K co-prime pairs of factors of a given number

Given two integers N and K, the task is to find K pair of factors of the number N such that the GCD of each pair of factors is 1. 
Note: K co-prime factors always exist for the given number
Examples:

Input: N = 6, K = 1 
Output: 2 3 
Explanation:
Since 2 and 3 are both factors of 6 and gcd(2, 3) = 1.
Input: N = 120, K = 4 
Output:
2 3 
3 4 
3 5 
4 5 

Naive Approach:
The simplest approach would be to check all the numbers upto N and check if the GCD of the pair is 1. 
Time Complexity: O(N²) 
Space Complexity: O(1)
Linear Approach:
Find all possible divisors of N and store in another array. Traverse through the array to search for all possible coprime pairs from the array and print them. 
Time Complexity: O(N) 
Space Complexity: O(N)
Efficient Approach:
Follow the steps below to solve the problem:

• It can be observed that if GCD of any number, say x, with 1 is always 1, i.e. GCD(1, x) = 1.
• Since 1 will always be a factor of N, simply print any K factors of N with 1 as the coprime pairs.

Below is the implementation of the above approach.

C++

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function prints the
// required pairs
void FindPairs(int n, int k)
{
    // First co-prime pair
    cout << 1 << " " << n << endl;

    // As a pair (1 n) has
    // already been Printed
    k--;

    for (long long i = 2;
         i <= sqrt(n); i++) {

        // If i is a factor of N
        if (n % i == 0) {

            cout << 1 << " "
                 << i << endl;
            k--;
            if (k == 0)
                break;

            // Since (i, i) won't form
            // a coprime pair
            if (i != n / i) {
                cout << 1 << " "
                     << n / i << endl;
                k--;
            }
            if (k == 0)
                break;
        }
    }
}

// Driver Code
int main()
{

    int N = 100;
    int K = 5;

    FindPairs(N, K);

    return 0;
}

Java

// Java implementation of
// the above approach
import java.util.*;

class GFG{

// Function prints the
// required pairs
static void FindPairs(int n, int k)
{

    // First co-prime pair
    System.out.print(1 + " " + n + "\n");

    // As a pair (1 n) has
    // already been Printed
    k--;

    for(long i = 2; i <= Math.sqrt(n); i++)
    {

        // If i is a factor of N
        if (n % i == 0)
        {
            System.out.print(1 + " " + i + "\n");
            k--;
            if (k == 0)
                break;

            // Since (i, i) won't form
            // a coprime pair
            if (i != n / i)
            {
                System.out.print(1 + " " +
                             n / i + "\n");
                k--;
            }
            if (k == 0)
                break;
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 100;
    int K = 5;

    FindPairs(N, K);
}
}

// This code is contributed by princiraj1992

Python3

# Python3 implementation of
# the above approach
from math import sqrt

# Function prints the
# required pairs
def FindPairs(n, k):

    # First co-prime pair
    print(1, n)

    # As a pair (1 n) has
    # already been Printed
    k -= 1

    for i in range(2, int(sqrt(n)) + 1):

        # If i is a factor of N
        if(n % i == 0):
            print(1, i)

            k -= 1
            if(k == 0):
                break

            # Since (i, i) won't form
            # a coprime pair
            if(i != n // i):
                print(1, n // i)
                k -= 1

            if(k == 0):
                break

# Driver Code
if __name__ == '__main__':

    N = 100
    K = 5

    FindPairs(N, K)

# This code is contributed by Shivam Singh

C#

// C# implementation of
// the above approach
using System;

class GFG{

// Function prints the
// required pairs
static void FindPairs(int n, int k)
{

    // First co-prime pair
    Console.Write(1 + " " + n + "\n");

    // As a pair (1 n) has
    // already been Printed
    k--;

    for(long i = 2; i <= Math.Sqrt(n); i++)
    {

        // If i is a factor of N
        if (n % i == 0)
        {
            Console.Write(1 + " " + i + "\n");
            k--;
            if (k == 0)
                break;

            // Since (i, i) won't form
            // a coprime pair
            if (i != n / i)
            {
                Console.Write(1 + " " +
                          n / i + "\n");
                k--;
            }
            if (k == 0)
                break;
        }
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 100;
    int K = 5;

    FindPairs(N, K);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript implementation of
// the above approach


// Function prints the
// required pairs
function FindPairs(n, k)
{
    // First co-prime pair
    document.write(1 + " " + n + "<br>");

    // As a pair (1 n) has
    // already been Printed
    k--;

    for (let i = 2;
        i <= Math.sqrt(n); i++) {

        // If i is a factor of N
        if (n % i == 0) {

            document.write( 1 + " "
                + i + "<br>");
            k--;
            if (k == 0)
                break;

            // Since (i, i) won't form
            // a coprime pair
            if (i != n / i) {
                document.write(1 + " "
                    + n / i + "<br>");
                k--;
            }
            if (k == 0)
                break;
        }
    }
}

// Driver Code

let N = 100;
let K = 5;

FindPairs(N, K);
// This code is contributed by _saurabh_jaiswal
</script>

Output:

1 100
1 2
1 50
1 4
1 25

Time Complexity: O(sqrt(N)) 
Auxiliary Space: O(1)