**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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