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 (Clang 10.0.0) (judge verdict: Accepted)

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

int f(int x, int y, int z)
{
	int r;
	r = x*x + y*y + z*z + x*y + y*z + z*x;
	return (r);
}
int	main(void)
{
	int n;
	scanf("%d",&n);
	int *arr;
	arr = calloc(sizeof(int),n);
	int i = 1;
	int j = 1;
	int k = 1;
	while (i < 99)
	{
		j = 1;
		while (j < 99)
		{
			k = 1;
			while (k < 99)
			{
				int index = f(i,j,k);
				if (i + j + k <= 140)
					arr[index-1]++;
				k++;
			}
			j++;
		}
		i++;
	}
	i = 0;
	while (i < n)
	{
		printf("%d\n",arr[i]);
		i++;
	}
	return (0);
}