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++

`// 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;` `}` |

## 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

`<?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.` `?>` |

## Javascript

`<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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