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>
using namespace std;
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
#define REPR(i, n) for (int i = (int)(n - 1); i >= 0; --i)
#define FOR(i, x, n) for (int i = x; i < (int)(n); ++i)
#define ALL(n) begin(n), end(n)
#define ll long long
#define INF 1e9

int ansfunc(int x, int y, int z) {
    return (x + y + z) * (x + y + z) - (x * y + y * z + z * x);
}

int main() {
    int N;
    cin >> N;
    vector<int> ans(N);
    REP(n, N) {
        FOR(i, 1, 100) {
            if (ansfunc(i, 1, 1) > n) continue;
            FOR(j, 1, i + 1) {
                if (ansfunc(i, j, 1) > n) continue;
                FOR(k, 1, j + 1) {
                    if (ansfunc(i, j, k) > n) continue;
                    if (ansfunc(i, j, k) == n) {
                        if (i == j && j == k && k == i)
                            ans[n]++;
                        else if (i != j && j != k && k != i)
                            ans[n] += 6;
                        else
                            ans[n] += 3;
                    }
                }
            }
        }
    }
    REP(i, N) cout << ans[i] << endl;
    return 0;
}