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 <stdio.h>

int calc(int x, int y, int z)
{
    int ans;
    
    ans = x*x + y*y + z*z + x*y + y*z + z*x;
    
    if(ans <= 10000)
      	return (ans);
    else
        return (0);
}

int	main(void)
{
  	int	n;
  	int x;
  	int y;
  	int z;
  	int	a[10002];
  	int	i;
  	int max;
  	int ans;
  	
  	max = 1000;
  	
  	scanf("%d", &n);
  
  	for(i=0; i<10001; i++)
  	{
  		a[i] = 0;
  	}
  
  	for(x=1; x<max; x++)
    {
        for(y=1; y<max; y++)
        {
            for(z=1; z<max; z++)
            {
                ans = calc(x,y,z);
                a[ ans ]++;
            }
        }
    }
  
    for(i=1; i<=n; i++)
      	printf("%d\n", a[i]);
  	
  	return (0);
}