# Represent K^N as the sum of exactly N numbers

Given two numbers N and K, the task is to represent KN as the sum of exactly N numbers. Print NA if no such numbers are possible.

Examples:

Input: N = 5, K = 2
Output: 2 2 4 8 16
Explanation:
2 + 2 + 4 + 8 + 16 = 32 = 25

Input: N = 4, K = 3
Output: 3 6 18 54
Explanation:
3 + 6 + 18 + 54 = 81 = 34

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to obtain numbers such that their sum is a power of K, we can choose the numbers that follow the condition:

``` ```

This will always give the sum as a power of K.

For example: This can be illustrated as:

```Let N = 3 and K = 4.

We need to represent 43 (=64)
as the sum of exactly 3 numbers

According to the mentioned approach,
The 3 numbers which can be chosen are
(41) = 4
(42 - 41) = 16 - 4 = 12
(43 - 42) = 64 - 16 = 48

Adding the numbers = 4 + 12 + 48 = 64
which is clearly 43

Therefore the required 3 numbers
are 4, 12 and 48.
```

Below is the implementation of the above approach:

## C++

 `// C++ program to represent K^N ` `// as the sum of exactly N numbers ` ` `  `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to print N numbers whose ` `// sum is a power of K ` `void` `print(ll n, ll k) ` `{ ` `    ``// Printing K ^ 1 ` `    ``cout << k << ``" "``; ` ` `  `    ``// Loop to print the difference of ` `    ``// powers from K ^ 2 ` `    ``for` `(``int` `i = 2; i <= n; i++) { ` `        ``ll x = ``pow``(k, i) - ``pow``(k, i - 1); ` `        ``cout << x << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``ll N = 3, K = 4; ` `    ``print(N, K); ` `    ``return` `0; ` `} `

## Java

 `// Java program to represent K^N ` `// as the sum of exactly N numbers ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to print N numbers whose ` `// sum is a power of K ` `static` `void` `print(``int` `n, ``int` `k) ` `{ ` `    ``// Printing K ^ 1 ` `    ``System.out.print(k+ ``" "``); ` `  `  `    ``// Loop to print the difference of ` `    ``// powers from K ^ 2 ` `    ``for` `(``int` `i = ``2``; i <= n; i++) { ` `        ``int` `x = (``int``) (Math.pow(k, i) - Math.pow(k, i - ``1``)); ` `        ``System.out.print(x+ ``" "``); ` `    ``} ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``3``, K = ``4``; ` `    ``print(N, K); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python 3

 `# Python 3 program to represent K^N ` `# as the sum of exactly N numbers ` `from` `math ``import` `pow` ` `  `# Function to print N numbers whose ` `# sum is a power of K ` `def` `printf(n, k): ` `     `  `    ``# Printing K ^ 1 ` `    ``print``(``int``(k),end ``=` `" "``) ` ` `  `    ``# Loop to print the difference of ` `    ``# powers from K ^ 2 ` `    ``for` `i ``in` `range``(``2``, n ``+` `1``, ``1``): ` `        ``x ``=` `pow``(k, i) ``-` `pow``(k, i ``-` `1``) ` `        ``print``(``int``(x),end``=` `" "``) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``N ``=` `3` `    ``K ``=` `4` `    ``printf(N, K) ` ` `  `# This code is contributed by Surendra_Gangwar `

## C#

 `// C# program to represent K^N ` `// as the sum of exactly N numbers ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to print N numbers whose ` `// sum is a power of K ` `static` `void` `print(``int` `n, ``int` `k) ` `{ ` `    ``// Printing K ^ 1 ` `    ``Console.Write(k+ ``" "``); ` `   `  `    ``// Loop to print the difference of ` `    ``// powers from K ^ 2 ` `    ``for` `(``int` `i = 2; i <= n; i++) { ` `        ``int` `x = (``int``) (Math.Pow(k, i) - Math.Pow(k, i - 1)); ` `        ``Console.Write(x+ ``" "``); ` `    ``} ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 3, K = 4; ` `    ``print(N, K); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```4 12 48
```

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