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>
#include <map>
using namespace std;

int main()
{
	long long n, res;
	cin >> n;
	map<long long, long long>count;
	for (long long x = 1; x <= n; ++x)
	{
		for (long long y = 1; y <= n - x; ++y)
		{
			for (long long z = 1; z <= n - y; ++z)
			{
				res = x * x + y * y + z * z + x * y + y * z + z * x;
				if (res <= n)
				{
					count[res]++;
				}
				else break;
			}
		}
	}
	for (long long i = 1; i <= n; ++i) cout << count[i] << endl;
	ios::sync_with_stdio(false);
	cin.tie();
	cout.tie();
	return 0;
}