Problem:
Score : 300 points

Problem Statement

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:

1 \leq x,y,z

x^2 + y^2 + z^2 + xy + yz + zx = n

Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Constraints

All values in input are integers.

1 \leq N \leq 10^4

Input

Input is given from Standard Input in the following format:

N

Output

Print N lines. The i-th line should contain the value f(i).

Sample Input 1

20

Sample Output 1

0
0
0
0
0
1
0
0
0
0
3
0
0
0
0
0
3
3
0
0

For n=6, only (1,1,1) satisfies both of the conditions. Thus, f(6) = 1.

For n=11, three triples, (1,1,2), (1,2,1), and (2,1,1), satisfy both of the conditions. Thus, f(6) = 3.

For n=17, three triples, (1,2,2), (2,1,2), and (2,2,1), satisfy both of the conditions. Thus, f(17) = 3.

For n=18, three triples, (1,1,3), (1,3,1), and (3,1,1), satisfy both of the conditions. Thus, f(18) = 3.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int nn;
  cin>>nn;
  for(int n=1;n<=nn;n++){
    int ans=0;
    int mx = sqrt(n/6);
    for(int x=1;x<=mx;x++){
      int my= sqrt(n/3 - x*x) ;
      for(int y=x;y<=my;y++){
        int mz= sqrt(n-5*x*x);
        for(int z=y;z<=mz;z++){
          if (x*x+y*y+z*z+x*y+y*z+z*x==n){
            if (x==y && y==z){
              ans=ans+1;
            }else if(x==y || y==z){
              ans=ans+3;
            }else{
              ans=ans+6;
            }     
          }
        }
      }
    }
    cout<<  ans<< endl;
  }
  
  
  
}