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<math.h>

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

  int max;
  int count;
  for(int i=1;i<=N;i++){
    count = 0;
    max = sqrt(N);
    for(int x=1;x<max;x++){
      for(int y=1;y<x+1;y++){
	for(int z=1;z<y+1;z++){
	  if(pow(x,2.0)+pow(y,2.0)+pow(z,2.0)+x*y+y*z+z*x == i){
	    if(x==y && y==z) count++;
	    else if(x==y || y==z || z==x) count += 3;
	    else count += 6;
	  }
	}
      }
    }
    printf("%d\n", count);
  }

  return 0;
}