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: Time Limit Exceeded)

#include <stdio.h>

int f[10001] = { 0 };
int n;
void dfs(int x, int y, int z) {
  int ret = x*x + y*y + z*z + x*y + y*z + z*x;
  if(ret > n) {
    return;
  }
  // fflush(stdin);
  // printf("%d %d %d\n", x, y, z);
  if(x == y && y == z) { f[ret] += 1; }
  else if(x == y || y == z || z == x) { f[ret] += 3; }
  else f[ret] += 6;
  dfs(x, y, z + 1);
  dfs(x, y+1, z+1);
  dfs(x+1, y+1, z+1);
}

int main(void) {
  scanf("%d", &n);
  dfs(1,1,1);
  for(int i = 1;i <= n;i++) {
    printf("%d\n", f[i]);
  }
  return 0;
}