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<math.h>
int main(){
 	int i;
  int n;
  int x,y,z;
  	int ans = 0;
  
  scanf("%d",&n);
  for(i=1;i<=n;i++){
 	for(x=1;x<=sqrt(i);x++){
      for(y=x;y<=sqrt(i);y++){
        for(z=y;z<=sqrt(i);z++){
      		if(x*x+y*y+z*z+x*y+y*z+z*x == i){
             if(x!=y&&y!=z&&z!=x){
               ans+=6;
             }
              else if(x==y&&y!=z){
                ans+=3;
              
              }
              else if(x!=y&&y==z){
               ans+=3; 
              }
              else if(x==y&&y==z){
               ans+=1; 
              }
              
              
            }
        }
      }
    }
    printf("%d\n",ans);
    ans=0;
  }

  return 0;
}