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 rep(i,n) for(int i=0;i<(int)n;++i)
#define REP(i,a,b) for(int i=int(a);i<(int)b;++i)
#define Graph vector<vecctor<ll>>
#define pairs vector<pair<ll,ll>>
#define pb push_back



using namespace std;
using ll = long long;

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}





int main(void)
{
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    vector<ll>ans(n);
    for(int k=1;k<=n;k++){
        if(k%4 == 0){
            ans[k-1] = ans[k/4-1];
            continue;
        }
    for(int i=2; i*i <= 2*k;i++){
        int les = 2*k-i*i;
        for(int j=2;j*j < les;j++){
            int les2 = les-j*j;
            for(int t=2;t*t <= les2;t++){
                if(t*t == les2){
                    if(i+j-t > 0 && i+t-j > 0 && j+t-i > 0) ans[k-1]++;
                }
            }
        }

    }
    cout << ans[k-1] << endl;
    }

    return 0;
   
}