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<stdio.h>
#include<math.h>
#define rep(i,l,m) for(int i=l;i<=(int)(m);i++)

int g(int n,int j){
  int zz=sqrt(n);
  return zz+j;
}

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

int main(){
  int N;
  int x,y,z,j;
  int start=1;
  int end=start+10;
  int G=0,count=0;
  if(scanf("%d",&N)>=1){
    rep(i,1,N){
      count=0;
      rep(j,start,end){
	G=g(i,j);
	rep(x,1,G-2){
	  rep(y,1,G-x-1){
	    z=G-x-y;
	    if(f(x,y,z)==i){
	      count++;
	      //printf("#%d %d %d\n",x,y,z);
	    }
	  }
	}

      }
      //printf("%d %d %d %d %d\n",i,count,G,start,end);
      //printf("%d %d\n",i,count);
      printf("%d\n",count);
      if(count!=0){
	start=j-3;
	end=start+10;
      }
    }
  }
  return 0;
}