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)

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

int main()
{
    int n,x,y,z,s,i;
    s=0;
    scanf("%d",&n);
   if(n==0)
   {
     printf("0");
   }
  for(i=1;i<=n;i++)
  {
    if((i==1 || i==2 || i==3 || i==4 || i==5) && i !=n)
    {
        printf("0\n");
    }
    else if(i >= 6)
    {
     for(x=1;x<=n-5;x++)
     {
         for(y=1;y<=n-5;y++)  
        {
            for(z=1;z<=n-5;z++)
            {
               if(x*x+y*y+z*z+x*y+y*z+x*z==n)
               {
                  s++;
                 break;
               }
            }
        }
     }
    }
    
    if(i == n)
    {
      printf("%d",s);
    }
    else
    {
      printf("%d\n",s);
      s=0;
    }
  }
return 0;
}