Pair of integers with difference K having an element as the K-th multiple of the other

Given an integer K, the task is to find a pair of numbers (A, B) such that A – B = K and A / B = K. If no such pair can be generated, print “No”.

Examples:

Input: K = 6 
Output: 7.2 1.2 
Explanation: 
Since 7.2 – 1.2 = 6 and 7.2 / 1.2 = 6, The pair {7.2, 1.2} satisfy the necessary conditions.

Input: K = 1 
Output: No 

Approach:
The following observations are needed to be made to solve the problem: 

• The given conditions can be written in the form of equations as: 
  1. Equation (1): A – B = K => A – B – K = 0
  2. Equation (2): A / B = K => A – (K * B) = 0
• Solving both these equations, we obtain:

( K * Equation (1) ) – Equation(2) = 0 
=> K*A – K*B – K² – (A – K*B) = 0 
=> K*A – K² – A – K*B + K*B = 0 
=> A*(K-1) – K² = 0 
=> A*(K-1) = K²
=> A = K² / (K -1)

• Replacing the value of A in Equation (1), the value of B can be obtained to be K / (K – 1)
• It can be observed that if the value of K is 1, then no such pair can be found as the denominator of both A and B becomes 0.

Follow the steps below to solve the problem: 

• If K is equal to 1, then print “NO”.
• Otherwise, if K is equal to any value other than 0, compute the values (K*K) / (K -1) and K / (K – 1) and print them.

Below is the implementation of the above approach: 

C++

// C++ Program to implement
// the above problem
#include <bits/stdc++.h>
using namespace std;

// Function to find the
// required pair
void computePair(double K)
{
    // No pair possible
    if (K == 1) {
        cout << "No";
        return;
    }
    else {

        cout << K * K / (K - 1) << " ";
        cout << K / (K - 1) << endl;
    }
}

// Driver Code
int main()
{
    double K = 6;
    computePair(K);

    return 0;
}

Java

// Java program to implement
// the above problem
class GFG{

// Function to find the
// required pair
static void computePair(double K)
{

    // No pair possible
    if (K == 1)
    {
        System.out.print("No");
        return;
    }
    else
    {
        System.out.print(K * K / (K - 1) + " ");
        System.out.print(K / (K - 1) + "\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    double K = 6;

    computePair(K);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above problem

# Function to find the
# required pair
def computePair(K):

    # No pair possible
    if (K == 1):
        print("No")
        return

    else:
        print(K * K / (K - 1), end = " ")
        print(K / (K - 1))

# Driver Code
if __name__ == "__main__":

    K = 6

    computePair(K)

# This code is contributed by chitranayal

C#

// C# program to implement
// the above problem
using System;

class GFG{

// Function to find the
// required pair
static void computePair(double K)
{

    // No pair possible
    if (K == 1)
    {
        Console.Write("No");
        return;
    }
    else
    {
        Console.Write(K * K / (K - 1) + " ");
        Console.Write(K / (K - 1) + "\n");
    }
}

// Driver Code
public static void Main(String[] args)
{
    double K = 6;

    computePair(K);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript Program to implement
// the above problem

// Function to find the
// required pair
function computePair(K)
{
    // No pair possible
    if (K == 1) {
        document.write( "No");
        return;
    }
    else {

        document.write( K * K / (K - 1) + " ");
        document.write( K / (K - 1) );
    }
}

// Driver Code
var K = 6;
computePair(K);

// This code is contributed by rutvik_56.
</script>

Output:

7.2 1.2

Time Complexity: O(1)
Auxiliary Space; O(1)