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 F first
#define S second
#define MP make_pair
#define pb push_back
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define LCM(a, b) (a) / __gcd((a), (b)) * (b)
#define CEIL(a, b) (a)/(b)+(((a)%(b))?1:0)
#define ln '\n'

using namespace std;
using LL = long long;
using ldouble = long double;
using P = pair<int, int>;
using LP = pair<LL, LL>;

static const int INF = INT_MAX;
static const LL LINF = LLONG_MAX;
static const int MIN = INT_MIN;
static const LL LMIN = LLONG_MIN;
static const int MOD = 1e9 + 7;
static const int SIZE = 200005;

const int dx[] = {0, -1, 1, 0};
const int dy[] = {-1, 0, 0, 1};

vector<LL> Div(LL n) {
    vector<LL> ret;
    for(LL i = 1; i * i <= n; ++i) {
        if(n % i == 0) {
            ret.pb(i);
            if(i * i != n) ret.pb(n / i);
        }
    }
    sort(all(ret));
    return ret;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int N;
    cin >> N;
    for(int i = 1; i <= N; ++i) {
        int cnt = 0;
        for(int x = 1; x <= 100; ++x) {
            for(int y = 1; x * x + y * y + x * y < i; ++y) {
                for(int z = 1; x * x + y * y + z * z + x * y + x * z + y * z <= i; ++z) {
                    if(x * x + y * y + z * z + x * y + x * z + y * z == i) {
                        ++cnt;
                    }
                }
            }
        }
        cout << cnt << endl;
    }
    return 0;
}