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 <bits/stdc++.h>
 
#define pii pair<int, int>
#define endl "\n"
#define int long long
#define F first
#define S second
#define B begin
#define E end
#define RB rbegin
#define RE rend
 
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
 
const int INF = 1e9 + 1;
const double EPS = 1e-10;
const ll P = 1e9 + 7;
const int Q = 47;
 
using namespace std;
 
void hello()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
}
 
void show(vector<int>& arr)
{
    for (auto i : arr)
        cout << i << endl;
    cout << endl;
}
void show(vector<vector<int>>& arr)
{
    for (auto i : arr)
        show(i);
    cout << endl;
}


signed main()
{
    hello();
    
    int n;
    cin >> n;
    vector <int> res(n);
    for (int i = 1; i <= 100; ++i)
        for (int j = 1; j <= 100; ++j)
            for (int k = 1; k <= 100; ++k)
                if (i * i + j * j + k * k + i * j + j * k + i * k <= n)
                    res[i * i + j * j + k * k + i * j + j * k + i * k - 1]++;
    show(res);
}