Count ways to express a number as sum of powers - GeeksforGeeks

26 Mar, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int x = 100, n = 2;    cout << countWays(x, n);    return 0;}
// 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
# 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 codex = 100n = 2 print(countWays(x, n)) # This code is contributed# by Anant Agarwal.
// 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.
<?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.?>
<script> // JavaScript 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    let val = (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.function countWays(x, n){    return countWaysUtil(x, n, 1);} // Driver code     let x = 100, n = 2;    document.write(countWays(x, n));     // This code is contributed by Mayank Tyagi </script>
Output: 
3
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