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 = 6Output:7.2 1.2Explanation:

Since 7.2 – 1.2 = 6 and 7.2 / 1.2 = 6, The pair {7.2, 1.2} satisfy the necessary conditions.

Input:K = 1Output: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:
- Equation (1):
**A – B = K => A – B – K = 0** - Equation (2):
**A / B = K => A – (K * B) = 0**

- Equation (1):
- Solving both these equations, we obtain:

( K * Equation (1) ) – Equation(2) = 0

=> K*A – K*B – K^{2 }– (A – K*B) = 0

=> K*A – K^{2}– A – K*B + K*B = 0

=> A*(K-1) – K^{2 }= 0

=> A*(K-1) = K^{2}

=>A = K^{2}/ (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 theof both**denominator****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` |

**Output:**

7.2 1.2

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

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