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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main(void){
  int N,n,nn,i,j,k,s;
  scanf("%d",&N);
  if(N<0||N<10000)
    return 0;
  
  for(n=1;n<=N;n++){
    s=0;
    for(i=1;i<=n/6;i++){
      for(j=1;j<=n/6;j++){
        for(k=1;k<=n/6;k++){
          nn=i*i+j*j+k*k+i*j+j*k+k*i;
          if(n==nn){
            s++;
          }
        }
      }
      printf("%d",s);
    }
  }
  return 0;
}