将K ^ N表示为正好N个数字的总和

给定两个数字N和K ，任务是将K ^N表示为正好N个数字的总和。如果没有这样的数字，则打印NA 。
例子：

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 
 

方法：为了获得数字，使它们的和为K的幂，我们可以选择满足以下条件的数字：

$i^{th} number = K^{i} - K^{i - 1}$

这将始终使总和为K的幂。
例如：这可以说明为：

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.

下面是上述方法的实现：

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

Javascript

 
// Javascript program to represent K^N
// as the sum of exactly N numbers
 
// Function to print N numbers whose
// sum is a power of K
function print(n, k)
{
    // Printing K ^ 1
    document.write( k + " ");
 
    // Loop to print the difference of
    // powers from K ^ 2
    for (var i = 2; i <= n; i++) {
        var x = Math.pow(k, i) - Math.pow(k, i - 1);
        document.write( x + " ");
    }
}
 
// Driver code
var N = 3, K = 4;
print(N, K);
 
// This code is contributed by rutvik_56.

输出：

4 12 48