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>
#include <math.h>

int g(int x, int y, int z) {
    return x * x + y * y + z * z + x * y + y * z + z * x;
}

int h(int x, int y, int z) {
    return x == y ? (y == z ? 1 : 3) : (y == z ? 3 : 6);
}

int main(void) {
    int a[10001] = {0};
    int n;
    scanf("%d", &n);

    for (int x = 1; ; x++) {
        int k = 6 * x * x;
        if (n < k) break;
        for (int y = x; ; y++) {
            k = x * x + 3 * y * y + 2 * x * y;
            if (n < k) break;
            for (int z = y; ; z++) {
                k = g(x, y, z);
                if (n < k) break;
                int w = h(x, y, z);
                a[k] += w;
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        printf("%d\n", a[i]);
    }
    return 0;
}