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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] ans = new int[10000];
		for(int x = 1; x < 100; x++){
			for(int y = 1; y < 100; y++){
				for(int z = 1; z < 100; z++){
					if((x* x)+(y*y)+(z*z)+(x*y)+(y*z)+(x*z)-1 < 10000)
						ans[(x* x)+(y*y)+(z*z)+(x*y)+(y*z)+(x*z)-1]++;
				}
			}
		}
		for(int i = 0; i< n - 1;i++){
			System.out.println(ans[i]);
		}
	}

	public static boolean isInteger(double number){
		return Math.ceil(number) == Math.floor(number);
	}

}