# The Power Sum

### Problem Statement :

```Find the number of ways that a given integer, X , can be expressed as the sum of the  Nth powers of unique, natural numbers.

For example, if X =  13 and N =  2 , we have to find all combinations of unique squares adding up to 13. The only solution is 2^2 + 3^2.

Function Description

Complete the powerSum function in the editor below. It should return an integer that represents the number of possible combinations.

powerSum has the following parameter(s):

X: the integer to sum to
N: the integer power to raise numbers to

Input Format

The first line contains an integer X.
The second line contains an integer N.

Constraints

1  <=   X  <=  1000
2  <=   N   <=   10

Output Format

Output a single integer, the number of possible combinations caclulated.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include <iostream>
#include <vector>

using namespace std;

int power (int a, int n) {
if(n == 0)
return 1;
// else
if(n % 2 == 0) {
int temp = power(a, n / 2);
return temp * temp;
}
// else
return a * power(a, n - 1);
}

int solve(int x, const vector<int> &powers, int index) {
if(index == 0) {
return (x == 1) ? 1 : 0;
}
// else
if(x == powers[index])
return 1 + solve(x, powers, index - 1);
// else
int res = 0;
res += solve(x - powers[index], powers, index - 1);
res += solve(x, powers, index - 1);
return res;
}

int main() {
int x, n;
cin >> x >> n;

int pow = 1;
vector<int> powers;
for(int a = 2; pow <= x; a++) {
powers.push_back(pow);
pow = power(a, n);
}

cout << solve(x, powers, powers.size() - 1) << endl;
return 0;
}

In     Java  :

import java.util.Scanner;

public class PowerSum {
int count=0;
int sum;
int pow;
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int x=in.nextInt();
int n=in.nextInt();
PowerSum p=new PowerSum();
p.sum=x;
p.pow=n;
int N=(int)Math.pow(x, (1.0/n));
p.getcount(p.sum,N,true);
p.getcount(p.sum,N,false);
System.out.println(p.count);
in.close();
}
void getcount(int sum1,int n,boolean lenyani){

if(lenyani==true){
sum1=sum1-(int)Math.pow(n, pow);

}
if(sum1<0) return;
if(sum1==0){
count++;
return;
}
if(n==1) return;
getcount(sum1,n-1,true);
getcount(sum1,n-1,false);
}

}

In    C   :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int the_power_sum(int n, int m,int p){
int tmp = pow(m,p);
if(tmp == n) return 1;
if(tmp > n) return 0;
return the_power_sum(n,m+1,p) + the_power_sum(n-tmp, m+1,p);
}
int main() {
int n,p;
scanf("%d%d",&n,&p);
printf("%d", the_power_sum(n,1,p));

return 0;
}

In   Python3 :

import sys
X = int(input())
N = int(input())

def rec(X,N,start):
count = 0
for i in range(start,X):
ans = X-i**N
if ans == 0:
count += 1
if ans> 0 :
count += rec(ans,N,i+1)
if ans<0:
continue
return count

print(rec(X,N,1))```
```

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