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 <iostream>
using namespace std;
#include <bits/stdc++.h>
#include<cmath>
int operation(int x, int y, int z)
{
    int val = (x*x)+(y*y)+(z*z)+(x*y)+(y*z)+(x*z);
    return val;
}
int main()
{
    int n;
    cin >>n;
    int counter=0;
    for(int i =1; i<=n; i++)
    {
        for(int x=1; x<100; x++)
        {
            for(int y=1; y<100; y++)
            {
                for(int z=1; z<100; z++)
                {
                    if(i == operation(x,y,z))
                    {
                        counter++;
                    }
                    else if(i < operation(x,y,z))
                        break;

                }
            }
        }

        cout<<counter<<endl;
        counter=0;
    }

    return 0;
}