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 <string.h>

int main(void) {
  int n, i;
  int a[10002] = {0};
  scanf("%d", &n);
  int x, y, z, aindex;

  for (x = 1; x <= n / 3; x++) {
    for (y = x; y <= n / 3; y++) {
      for (z = y; z <= n / 3; z++) {
        aindex = x * x + y * y + z * z + x * y + y * z + z * x;
        if (aindex < 10001) {
          if (x == y && y == z) {
            // printf("aindex:%d x=%d, y=%d, z=%d\n", aindex,x,y,z);
            a[aindex] += 1;
          } else if (x == y || y == z || z == x)
            a[aindex] += 3;
          else
            a[aindex] += 6;
        }
      }
    }
  }

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