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 <stdlib.h>
#include <math.h>

#define rep(i, n) for (int i = 0; i < n; i = i + 1)
#define MAX 10010
#define MAX_INT 9223372036854775807
#define array(N, t) (t *)calloc(N, sizeof(t))

int main() {
  int N;
  scanf("%d", &N);

  int * f = array(N + 1, int);

  rep(i, 100) {
    int x = i + 1;
    rep(j, 100) {
      int y = j + 1;
      rep(k, 100) {
        int z = k + 1;
        int sum = (int)(((int)pow(x + y, 2) +
                         (int)pow(x + z, 2) +
                         (int)pow(y + z, 2)) / 2);
        if (sum <= N) {
          f[sum] += 1;
        }
      }
    }
  }

  rep(i, N) {
    printf("%d\n", f[i + 1]);
  }

  free(f);
  return 0;
}