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: Wrong Answer)

#include<bits/stdc++.h>

#define ll long long
#define ull unsigned long long

#define pii pair < int, int >
#define pll pair < ll, ll >
#define pci pair < char, int >
#define psi pair < string, int >

#define vi vector < int >
#define vll vector < ll >
#define vs vector < string >
#define vc vector < char >
#define pvi vector < pii >
#define pvll vector < pll >

#define pb push_back

#define fast ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

using namespace std;

const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;

int main()
{
    fast;

    int n;

    cin >> n;

    int a[n + 5], tmp;

    memset(a, 0, sizeof(a));

    for(int i = 1; ; i++){
        for(int j = 1; ; j++){
            for(int k = 1; ; k++){
                tmp = (i * i) + (j * j) + (k * k) + (i * j) + (j * k) + (i * k);

                if(tmp > n) break;

                int cnt = 0;

                if(i == j && j == k) cnt = 1;
                else if(i == j || j == k) cnt = 3;
                else if(i != j && i != k && j != k) cnt = 6;

                a[tmp] = cnt;
            }

            int tmp1 = (i * i) + (j * j) + (i * j) + 1 + i + j;

            if(tmp1 > n) break;
        }

        int tmp2 = (i * i) + 3 + (2 * i);

        if(tmp2 > n) break;
    }

    for(int i = 1; i <= n; i++) cout << a[i] << endl;

    return 0;
}