# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

## C++

 `// C++ program to count number of ways ` `// to express x as sum of n-th power ` `// of unique natural numbers. ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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. `

## C#

 `// 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

 ` `

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Improved By : vt_m, jit_t

Article Tags :
Practice Tags :

6

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