Count ways to express a number as sum of powers

Given two integers x and n, we need to find number of ways to express x as sum of n-th powers of unique natural numbers. It is given that 1 <= n <= 20.

Examples:

Input  : x = 100
         n = 2
Output : 3
Explanation: There are three ways to 
express 100 as sum of natural numbers
raised to power 2.
100 = 10^2 = 8^2+6^2 = 1^2+3^2+4^2+5^2+7^2

Input  : x = 100
         n = 3
Output : 1
Explanation : The only combination is,
1^3 + 2^3 + 3^3 + 4^3

We use recursion to solve the problem. We first check one by one that the number is included in summation or not.

C++



filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count number of ways
// to express x as sum of n-th power
// of unique natural numbers.
#include <bits/stdc++.h>
using namespace std;
  
// num is current num.
int countWaysUtil(int x, int n, int num)
{
    // Base cases
    int val = (x - pow(num, n));
    if (val == 0)
        return 1;
    if (val < 0)
        return 0;
  
    // Consider two possibilities, num is
    // included and num is not included.
    return countWaysUtil(val, n, num + 1) +
           countWaysUtil(x, n, num + 1);
}
  
// Returns number of ways to express
// x as sum of n-th power of two.
int countWays(int x, int n)
{
    return countWaysUtil(x, n, 1);
}
  
// Driver code
int main()
{
    int x = 100, n = 2;
    cout << countWays(x, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count number of ways
// to express x as sum of n-th power
// of unique natural numbers.
public class GFG { 
  
    // num is current num.
    static int countWaysUtil(int x, int n, int num)
    {
        // Base cases
        int val = (int) (x - Math.pow(num, n));
        if (val == 0)
            return 1;
        if (val < 0)
            return 0;
       
        // Consider two possibilities, num is
        // included and num is not included.
        return countWaysUtil(val, n, num + 1) +
               countWaysUtil(x, n, num + 1);
    }
       
    // Returns number of ways to express
    // x as sum of n-th power of two.
    static int countWays(int x, int n)
    {
        return countWaysUtil(x, n, 1);
    }
       
    // Driver code
    public static void main(String args[])
    {
        int x = 100, n = 2;
        System.out.println(countWays(x, n));
    }
}
// This code is contributed by Sumit Ghosh

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to count number of ways
# to express x as sum of n-th power
# of unique natural numbers.
  
# num is current num.
def countWaysUtil(x,n,num):
  
    # Base cases
    val = (x - pow(num, n))
    if (val == 0):
        return 1
    if (val < 0):
        return 0
   
    # Consider two possibilities, num is
    # included and num is not included.
    return countWaysUtil(val, n, num + 1) +\
           countWaysUtil(x, n, num + 1)
  
   
# Returns number of ways to express
# x as sum of n-th power of two.
def countWays(x,n):
    return countWaysUtil(x, n, 1)
  
      
# Driver code
x = 100
n = 2
  
print(countWays(x, n))
  
# This code is contributed
# by Anant Agarwal.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to count number of ways
// to express x as sum of n-th power
// of unique natural numbers.
using System;
  
public class GFG { 
  
    // num is current num.
    static int countWaysUtil(int x,
                          int n, int num)
    {
          
        // Base cases
        int val = (int) (x - Math.Pow(num, n));
        if (val == 0)
            return 1;
        if (val < 0)
            return 0;
      
        // Consider two possibilities,
        // num is included and num is
        // not included.
        return countWaysUtil(val, n, num + 1)
              + countWaysUtil(x, n, num + 1);
    }
      
    // Returns number of ways to express
    // x as sum of n-th power of two.
    static int countWays(int x, int n)
    {
        return countWaysUtil(x, n, 1);
    }
      
    // Driver code
    public static void Main()
    {
        int x = 100, n = 2;
          
        Console.WriteLine(countWays(x, n));
    }
}
  
// This code is contributed by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count number of ways
// to express x as sum of n-th power
// of unique natural numbers.
  
// num is current num.
function countWaysUtil($x, $n, $num)
{
      
    // Base cases
    $val = ($x - pow($num, $n));
    if ($val == 0)
        return 1;
    if ($val < 0)
        return 0;
  
    // Consider two possibilities, num is
    // included and num is not included.
    return (countWaysUtil($val, $n, $num + 1) +
            countWaysUtil($x, $n, $num + 1));
}
  
// Returns number of ways to express
// x as sum of n-th power of two.
function countWays($x, $n)
{
    return countWaysUtil($x, $n, 1);
}
  
// Driver code
$x = 100; $n = 2;
echo(countWays($x, $n));
  
// This code is contributed by Ajit.
?>

chevron_right



Output:

3

This article is contributed by Anjali. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m, jit_t



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.