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;
#define rep(i,n) for (int i = 0; i < (n); i++) 
#define all(v) begin(v),end(v)
using ll = long long;

int main() {
  int n;cin>>n;
  vector<int>vec(n+1);
  for(int l=1;l<=n;l++){
    bool ans=false;
  for(int i=1;i<=100;i++){
    for(int j=1;j<=100;j++){
      for(int k=1;k<=100;k++){
        if(i*i+j*j+k*k+i*j+j*k+k*i==l&&i==j&&j==k&&k==i){
          vec[l]=1;
          ans=true;
        break;
        }
        else if(i*i+j*j+k*k+i*j+j*k+k*i==l&&i!=j&&j!=k&&k!=i){
          vec[l]=6;
          ans=true;
          break;
        }
        else if(i*i+j*j+k*k+i*j+j*k+k*i==l){
          vec[l]=3;
          ans=true;
          break;
        }
      }
      if(ans)break;
    }
    if(ans)break;
  }
    cout<<vec[l]<<endl;
  }
}