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>
using namespace std;

typedef long long ll;

ll n;

ll solve(ll x, ll y, ll z){
    return ((x*x+y*y+z*z)+(x*y+y*z+z*x));
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> n;
    ll arr[n+1];
    for(ll i=0; i<=n; i++) arr[i]=0;

    ll x;

    for(ll i=1; i<=130; i++){
        bool done=true;
        for(ll j=1; j<=130; j++){
            bool flag=true;
            for(ll k=1; k<=130; k++){
                x = solve(i,j,k);
                if(x<=n){
                    arr[x]++;
                    flag=false;
                    done=false;
                }
                if(x>n) break;
            }
            if(flag) break;
        }
        if(done) break;
    }
    for(int i=1; i<=n; i++) cout<<arr[i]<<'\n';

    return 0;
}