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<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
typedef long long ll;
typedef long l;
int return_max(int a,int b){
  if(a>b){
    return a;
  }else{
    return b;
  }
}

int return_min(int a,int b){
  if(a<b){
    return a;
  }else{
    return b;
  }
}


int main(void)
{

  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++){
    int count=0;
    for(int x=1;x<=(int)sqrt(n);x++){
      for(int y=x;y<=(int)sqrt(n-x*x);y++){
        for(int z=y;z<=(int)sqrt(n-x*x-y*y-x*y);z++){
          if(x*x+y*y+z*z+x*y+y*z+z*x==i){
            if(x==y&&y==z&&z==x){
              count++;
            }else{
              count+=3;
            }
          }
        }
      }
    }
    printf("%d\n",count);
  }

  return 0;
}