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<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<string>
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define reps(i,n) for(int i=1;i<=(int)n;i++)
#define all(v) v.begin(), v.end()
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define lb lower_bound
#define ub upper_bound
#define LB(a,x) lb(all(a), x) - a.begin()
#define UB(a,x) ub(all(a), x) - a.begin()
#define MOD 1000000007
#define itn int
#define enld endl
using namespace std;
typedef long long ll;
const double pi = 3.141592653589793;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return 1;} return 0;}
template<class T> bool chmin(T &a, const T &b){if(b<a){a=b; return 1;} return 0;}
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

int main(){
    int n; cin >> n;
    int cnt[10005];
    Fill(cnt, 0);
    for(int x=1;x<=100;x++){
        for(int y=1;y<=100-x;y++){
            for(int z=1;z<=100-x-y;z++){
                int sum = x*x+y*y+z*z+x*y+y*z+z*x;
                if(1 <= sum && sum <= n) cnt[sum]++;
            }
        }
    }
    reps(i,n){
        cout << cnt[i] << "\n";
    }
    return 0;
}