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 <math.h>

#define rep(i, n) for (int i = 0; i < n; i = i + 1)
#define MAX 10010
#define MAX_INT 9223372036854775807

int f(int n) {
  // pow(x, 2) + pow(y, 2) + pow(z, 2) + x * y + y * z + z * x == n;
  // (pow((x + y), 2) + pow((x + z), 2) + pow((y + z), 2)) / 2 == n;
  int ret_val = 0;
  for (int x = 1; x <= n; x += 1) {
    for (int y = 1; y <= n; y += 1) {
      for (int z = 1; z <= n; z += 1) {
        if ((int)pow(x, 2) + (int)pow(y, 2) + (int)pow(z, 2) + (x * y) + (y * z) + (z * x) == n) {
        //        if ((pow((x + y), 2) + pow((x + z), 2) + pow((y + z), 2)) / 2 == n) {
          ret_val += 1;
        }
      }
    }
  }
  return ret_val;
}

int main() {
  int N;

  scanf("%d", &N);

  for (int i = 1; i <= 20; i += 1) {
    printf("%d\n", f(i));
  }

  return 0;
}