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;

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

int main()
{
  int N,i =1;
  cin>>N;
  int x=1,y=1,z=1,i_max,test;
  unordered_map<int,int> m;
  while(test<N){
      test = equ(x,y,z);
      x++;
      i_max = x;
  }
  int limit = i_max*i_max*i_max;
  x=1;
  while(i<limit){
      int sum = equ(x,y,z);
      cout<<x<<y<<z<<" "<<sum<<endl;
      if(sum > N*N){
          break;
      }
      if(sum>N){
          if(y>= z && y!=1 && z!=1){
              x++;
              y=1;
              z=0;
          }
          else{
              y++;
              z=0;
          }
      }
      if(sum<N){
          m[sum]++;
          i--;
         
      }
      z++;
      i++;
  }
  for(int i=1;i<=N;i++){
      if(m.find(i)==m.end()){
          int temp=0;
          cout<<temp<<endl;
      }
      else{
          cout<<m[i]<<endl;
      }
  }
  
  return 0;
      
}