Find ways an Integer can be expressed as sum of n-th power of unique natural numbers

Given two numbers x and n, find number of ways x can be expressed as sum of n-th power of unique natural numbers.

Examples :

Input  : x = 10, n = 2
Output : 1
Explanation: 10 = 1² + 3², 
Hence total 1 possibility

Input  : x = 100, n = 2
Output : 3
Explanation: 100 = 10²
             OR 6² + 8²
             OR 1² + 3² + 4² + 5² + 7²
Hence total 3 possibilities

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is simple. We iterate through all number starting from 1. For every number, we recursively try all greater numbers and if we are able to find sum, we increment result.

C++

// C++ program to count number of ways any 
// given integer x can be expressed as n-th 
// power of unique natural numbers. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to calculate and return the 
// power of any given number 
int power(int num, unsigned int n) 
{ 
    if (n == 0) 
        return 1; 
    else if (n%2 == 0) 
        return power(num, n/2)*power(num, n/2); 
    else
        return num*power(num, n/2)*power(num, n/2); 
} 
  
// Function to check power representations recursively 
int checkRecursive(int x, int n, int curr_num = 1, 
                   int curr_sum = 0) 
{ 
    // Initialize number of ways to express 
    // x as n-th powers of different natural 
    // numbers 
    int results = 0; 
  
    // Calling power of 'i' raised to 'n' 
    int p = power(curr_num, n); 
    while (p + curr_sum < x) 
    { 
        // Recursively check all greater values of i 
        results += checkRecursive(x, n, curr_num+1, 
                                        p+curr_sum); 
        curr_num++; 
        p = power(curr_num, n); 
    } 
  
    // If sum of powers is equal to x 
    // then increase the value of result. 
    if (p + curr_sum == x) 
        results++; 
  
    // Return the final result 
    return results; 
} 
  
// Driver Code. 
int main() 
{ 
    int x = 10, n = 2; 
    cout << checkRecursive(x, n); 
    return 0; 
} 

Java

// Java program to count number of ways any  
// given integer x can be expressed as n-th  
// power of unique natural numbers.  
  
class GFG 
{ 
      
// Function to calculate and return the  
// power of any given number  
static int power(int num, int n)  
{  
    if (n == 0)  
        return 1;  
    else if (n % 2 == 0)  
        return power(num, n / 2) * power(num, n / 2);  
    else
        return num * power(num, n / 2) * power(num, n / 2);  
}  
  
// Function to check power representations recursively  
static int checkRecursive(int x, int n, int curr_num,int curr_sum)  
{  
    // Initialize number of ways to express  
    // x as n-th powers of different natural  
    // numbers  
    int results = 0;  
  
    // Calling power of 'i' raised to 'n'  
    int p = power(curr_num, n);  
    while (p + curr_sum < x)  
    {  
        // Recursively check all greater values of i  
        results += checkRecursive(x, n, curr_num + 1,  
                                        p + curr_sum);  
        curr_num++;  
        p = power(curr_num, n);  
    }  
  
    // If sum of powers is equal to x  
    // then increase the value of result.  
    if (p + curr_sum == x)  
        results++;  
  
    // Return the final result  
    return results;  
}  
  
// Driver Code. 
public static void main (String[] args)  
{ 
    int x = 10, n = 2;  
    System.out.println(checkRecursive(x, n, 1, 0)); 
}  
} 
  
// This code is contributed by mits 

Python3

# Python3 program to count number of ways any 
# given integer x can be expressed as n-th 
# power of unique natural numbers. 
  
# Function to calculate and return the 
# power of any given number 
def power(num, n): 
  
    if(n == 0): 
        return 1
    elif(n % 2 == 0): 
        return power(num, n // 2) * power(num, n // 2) 
    else: 
        return num * power(num, n // 2) * power(num, n // 2) 
  
# Function to check power representations recursively 
def checkRecursive(x, n, curr_num=1, curr_sum=0): 
  
    # Initialize number of ways to express 
    # x as n-th powers of different natural 
    # numbers 
    results = 0
  
    # Calling power of 'i' raised to 'n' 
    p = power(curr_num, n) 
    while(p + curr_sum < x): 
  
        # Recursively check all greater values of i 
        results += checkRecursive(x, n, curr_num + 1, p + curr_sum) 
        curr_num = curr_num + 1
        p = power(curr_num, n) 
  
    # If sum of powers is equal to x 
    # then increase the value of result. 
    if(p + curr_sum == x): 
        results = results + 1
  
    # Return the final result 
    return results 
  
# Driver Code. 
if __name__=='__main__': 
    x = 10
    n = 2
    print(checkRecursive(x, n)) 
  
  
# This code is contributed by 
# Sanjit_Prasad 

C#

// C# program to count number of ways any  
// given integer x can be expressed as  
// n-th power of unique natural numbers.  
using System; 
  
class GFG 
{ 
      
// Function to calculate and return  
// the power of any given number  
static int power(int num, int n)  
{  
    if (n == 0)  
        return 1;  
    else if (n % 2 == 0)  
        return power(num, n / 2) *  
               power(num, n / 2);  
    else
        return num * power(num, n / 2) * 
                     power(num, n / 2);  
}  
  
// Function to check power 
// representations recursively  
static int checkRecursive(int x, int n,  
                          int curr_num, 
                          int curr_sum)  
{  
    // Initialize number of ways to express  
    // x as n-th powers of different natural  
    // numbers  
    int results = 0;  
  
    // Calling power of 'i' raised to 'n'  
    int p = power(curr_num, n);  
    while (p + curr_sum < x)  
    {  
        // Recursively check all greater values of i  
        results += checkRecursive(x, n, curr_num + 1,  
                                        p + curr_sum);  
        curr_num++;  
        p = power(curr_num, n);  
    }  
  
    // If sum of powers is equal to x  
    // then increase the value of result.  
    if (p + curr_sum == x)  
        results++;  
  
    // Return the final result  
    return results;  
}  
  
// Driver Code. 
public static void Main()  
{ 
    int x = 10, n = 2;  
    System.Console.WriteLine(checkRecursive(x, n, 1, 0)); 
}  
} 
  
// This code is contributed by mits 

PHP

<?php 
// PHP program to count  
// number of ways any 
// given integer x can 
// be expressed as n-th 
// power of unique  
// natural numbers. 
  
// Function to calculate and return 
// the power of any given number 
function power($num, $n) 
{ 
      
    if ($n == 0) 
        return 1; 
    else if ($n % 2 == 0) 
        return power($num, (int)($n / 2)) *  
               power($num, (int)($n / 2)); 
    else
        return $num * power($num, (int)($n / 2)) *  
                      power($num, (int)($n / 2)); 
} 
  
// Function to check power 
// representations recursively 
function checkRecursive($x, $n, 
                        $curr_num = 1,  
                        $curr_sum = 0) 
{ 
      
    // Initialize number of 
    // ways to express 
    // x as n-th powers  
    // of different natural 
    // numbers 
    $results = 0; 
  
    // Calling power of 'i'  
    // raised to 'n' 
    $p = power($curr_num, $n); 
    while ($p + $curr_sum < $x) 
    { 
          
        // Recursively check all 
        // greater values of i 
        $results += checkRecursive($x, $n,  
                                   $curr_num + 1, 
                                   $p + $curr_sum); 
        $curr_num++; 
        $p = power($curr_num, $n); 
    } 
  
    // If sum of powers  
    // is equal to x 
    // then increase the 
    // value of result. 
    if ($p + $curr_sum == $x) 
        $results++; 
  
    // Return the final result 
    return $results; 
} 
  
// Driver Code. 
$x = 10; $n = 2; 
echo(checkRecursive($x, $n)); 
  
// This code is contributed by Ajit. 
?> 

Output:

Alternate Solution :

Below is an alternate simpler solution provided by Shivam Kanodia.

C++

// C++ program to find number of ways to express  
// a number as sum of n-th powers of numbers. 
#include<bits/stdc++.h> 
using namespace std; 
  
int res = 0; 
int checkRecursive(int num, int x, int k, int n) 
{ 
    if (x == 0)  
        res++; 
      
    int r = (int)floor(pow(num, 1.0 / n)); 
  
    for (int i = k + 1; i <= r; i++)  
    { 
        int a = x - (int)pow(i, n); 
        if (a >= 0) 
            checkRecursive(num, x -  
                          (int)pow(i, n), i, n); 
    } 
    return res; 
} 
  
// Wrapper over checkRecursive() 
int check(int x, int n) 
{ 
    return checkRecursive(x, x, 0, n); 
} 
  
// Driver Code 
int main() 
{  
    cout << (check(10, 2)); 
    return 0; 
} 
  
// This code is contributed by mits 

Java

// Java program to find number of ways to express a 
// number as sum of n-th powers of numbers. 
import java.io.*; 
import java.util.*; 
  
public class Solution { 
  
    static int res = 0; 
    static int checkRecursive(int num, int x, int k, int n) 
    { 
        if (x == 0)  
            res++; 
          
        int r = (int)Math.floor(Math.pow(num, 1.0 / n)); 
  
        for (int i = k + 1; i <= r; i++) { 
            int a = x - (int)Math.pow(i, n); 
          if (a >= 0) 
            checkRecursive(num,  
                   x - (int)Math.pow(i, n), i, n); 
        } 
        return res; 
    } 
      
    // Wrapper over checkRecursive() 
    static int check(int x, int n) 
    { 
        return checkRecursive(x, x, 0, n); 
    } 
  
    public static void main(String[] args) 
    {  
        System.out.println(check(10, 2)); 
    } 
} 

Python3

# Python 3 program to find number of ways to express  
# a number as sum of n-th powers of numbers. 
from math import pow, floor 
  
res = 0
def checkRecursive(num, x, k, n): 
    global res 
    if (x == 0): 
        res += 1
      
    r = floor(pow(num, (1 / n))); 
  
    for i in range(k + 1, r + 1, 1): 
        a = x - int(pow(i, n)) 
        if (a >= 0): 
            checkRecursive(num, x - int(pow(i, n)), i, n) 
    return res 
  
# Wrapper over checkRecursive() 
def check(x, n): 
    return checkRecursive(x, x, 0, n) 
  
# Driver Code 
if __name__ == '__main__': 
    print(check(10, 2)) 
          
# This code is contributed by 
# Surendra_Gangwar 

C#

// C# program to find number of 
// ways to express a number as sum 
// of n-th powers of numbers. 
using System; 
  
class Solution { 
  
    static int res = 0; 
    static int checkRecursive(int num, int x, 
                                int k, int n) 
    { 
        if (x == 0)  
            res++; 
          
        int r = (int)Math.Floor(Math.Pow(num, 1.0 / n)); 
  
        for (int i = k + 1; i <= r; i++)  
        { 
            int a = x - (int)Math.Pow(i, n); 
        if (a >= 0) 
            checkRecursive(num, x -  
                          (int)Math.Pow(i, n), i, n); 
        } 
        return res; 
    } 
      
    // Wrapper over checkRecursive() 
    static int check(int x, int n) 
    { 
        return checkRecursive(x, x, 0, n); 
    } 
      
    // Driver code 
    public static void Main() 
    {  
        Console.WriteLine(check(10, 2)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find number  
// of ways to express a number  
// as sum of n-th powers of numbers. 
$res = 0; 
  
function checkRecursive($num, $x,  
                        $k, $n) 
{ 
    global $res; 
    if ($x == 0)  
        $res++; 
      
    $r = (int)floor(pow($num,  
                        1.0 / $n)); 
  
    for ($i = $k + 1;  
         $i <= $r; $i++)  
    { 
        $a = $x - (int)pow($i, $n); 
        if ($a >= 0) 
            checkRecursive($num, $x -  
                    (int)pow($i, $n),  
                             $i, $n); 
    } 
    return $res; 
} 
  
// Wrapper over 
// checkRecursive() 
function check($x, $n) 
{ 
    return checkRecursive($x, $x,  
                          0, $n); 
} 
  
// Driver Code 
echo (check(10, 2)); 
  
// This code is contributed by ajit 
?> 

Output:

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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