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; ` `} ` |

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## 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 ` |

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## 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 ` |

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**Output:**

7.2 1.2

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

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