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 <string.h>
#include <math.h>

int main(void){
  int n;
  scanf("%d", &n);
  int a[n+1];
  int x, y, z;
  int i;
  for(i=1; i<n+1; i++){
    a[i]=0;
  }
  
  for(x=1; x<110; x++){
    for(y=1; y<110; y++){
      for(z=1; z<110; z++){
        i=0;
        i=x*x+y*y+z*z+x*y+y*z+z*x;
        if(i > n){
          continue;
        }else{
          a[i]+=1;
        }
      }
    }
  }
  
  for(i=1; i<n+1; i++){
    printf("%d\n", a[i]);
  }
  return 0;
}