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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main{
  public static void main(String[] args){
	  Scanner sc = new Scanner(System.in);
	  int n=sc.nextInt(),max=0;
	  for(int i=1;i<=n;i++) {
		  int count=0,f=0;
		  for(int x=1;x<=100;x++) {
			  for(int y=x;y<=100;y++) {
				  for(int z=y;z<=100;z++) {
					  f=x*x+y*y+z*z+x*y+y*z+z*x;
					  if(f==i) {
						  if(x!=y&&y!=z&&z!=x)count+=6;
						  else if(x==y&&y==z)count++;
						  else count+=3;
					  }
				  }
			  }
		  }
		  System.out.println(count);
	  }
  }
}