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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;


int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n;
  cin>>n;
  for(int i=1;i<=min(n,2);i++){
      cout<<0<<"\n";
  }
  for(int p=3;p<=n;p++){
    int y=1;
    int z=sqrt(p)-1;
    int sum=0;
    for(int i=y;i<=z;i++){
        for(int j=y;j<=z;j++){
          for(int k=y;k<=z;k++){
            if(i*i+j*j+k*k+i*j+i*k+j*k==p){
              if(i==j && j==k){
                sum+=1;
              }
              if((i==j && j!=k)||(j==k && i!=k)||(i==k && i!=j)){
                 sum+=3;
              }
              if(i!=j && j!=k && i!=k){
                 sum+=6;
              }
              
              i=max({i,j,k})+1;
                j=i;
                k=i;
            }
          }
        }
    }
    
     cout<<sum<<"\n";
    






  }




}