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: Wrong Answer)

// max_x = 98

#include <math.h>

int cor(int x, int y, int z, int n)
{
  if (n = x*x + y*y + z*z + x*y + y*z + z*x)
    return 1;
  return 0;
}
int main()
{
  int n;
  scanf("%d", &n);
  int x, y, z;
  int minx;
  int i, count;
//  int tx, ty, tz;
  for (i = 1; i <= n; i++)
  {
    count = 0;
    x = sqrt(i - 2) - 1;
//    y = sqrt(i - 3*x*x/4 - x/2 - 3/4) - (x+1)/2;
//    z = sqrt(i - x*x - y*y - x*y) - (x+y)/2;
    y = x;
    z = y;
    minx = sqrt(i / 6);
    while (x >= minx)
    {
      while (y >= 1)
      {
        z = y;
        while (z >= 1)
        {
          if (x*x + y*y + z*z + x*y + y*z + z*x == i)
          {
            if (x != y && y != z)
              count += 6;
            else if ((x != y && y == z) || (x == y && y != z))
              count += 3;
            else if (x == y && y == z)
              count++;
          }
          z--;
        }
        y--;
      }
      x--;
    }
/*    tx = x;
    ty = y;
    tz = z;
    if (cor(x, y, z, i) == 1)
      count++;
    while (y >= z)
    {
      y--;
    if (x >= 1 && y >= 1 && z >= 1)
    {
      if (x*x + y*y + z*z + x*y + y*z + z*x == i)
      {
        if (x != y && y != z)
          count += 6;
        else if ((x != y && y == z) || (x == y && y != z))
          count += 3;
        else if (x == y && y == z)
          count++;
      }
    }
//    }*/
    
    printf("%d\n", count);
  }
}