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: Time Limit Exceeded)

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

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