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

• Difficulty Level : Hard
• Last Updated : 16 Apr, 2021

Given two numbers x and n, find a 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

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 ``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`

## Python

 `# 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

 ``

## Javascript

 ``
Output
`1`

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``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.`  `def` `checkRecursive(num, rem_num, next_int, n, ans``=``0``):` `    ``if` `(rem_num ``=``=` `0``):``        ``ans ``+``=` `1` `    ``r ``=` `int``(num``*``*``(``1` `/` `n))` `    ``for` `i ``in` `range``(next_int ``+` `1``, r ``+` `1``):``        ``a ``=` `rem_num ``-` `int``(i``*``*``n)``        ``if` `a >``=` `0``:``            ``ans ``+``=` `checkRecursive(num, rem_num ``-` `int``(i``*``*``n), i, n, ``0``)``    ``return` `ans` `# 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

 `= 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``?>`

## Javascript

 ``
Output
`1`

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