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# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;

class Class1
{
    static void Main(string[] args)
    {
        int N = int.Parse(Console.ReadLine());

        for (int i = 1; i <= N; i++)
        {
            int result = 0;
            int leastNum = 1;
            int sqrt_i = (int) Math.Sqrt(i);
            for (int x = sqrt_i; x > 0; x--)
            {
                for (int y = sqrt_i; y > 0; y--)
                {
                    for (int z = leastNum; z <= sqrt_i; z++)
                    {
                        int n = (int) Math.Pow(x + y + z, 2) - (x * y + y * z + z * x);
                        if (n == i)
                        {
                            leastNum = z;
                            result++;
                            // xyzの組み合わせを簡単にする
                            if (z == y && z == x)
                            {
                                y = x = 0;
                                break;
                            }

                            // xyzの組み合わせを簡単にする
                            if ((z != y && z != x && y == x) ||
                                (z == y && z != x))
                            {
                                result += 2;
                                y = x = 0;
                                break;
                            }

                            // xyzの組み合わせを簡単にする
                            if ((z != y && z != x && x != y))
                            {
                                result += 5;
                                y = x = 0;
                                break;
                            }
                        }
                    }
                }
            }

            Console.WriteLine(result);
        }
    }
}