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>
int main(){
	int n;
  	scanf("%d",&n);
  	int i=1,x,y,z,sum,j=0;
  	for(i;i<=n;i++){
      for(x=1;x<=9;x++){
          for(y=x;y<=9;y++){
              for(z=y;z<=9;z++){
                sum=x*x+y*y+z*z+x*y+y*z+z*x;
                if(sum==i){
                  if((x==y)&&(y==z)&&(z==x))printf("1");
                  else if((x!=y)&&(y!=z)&&(z!=x))printf("6");
                  else printf("3");
                  j++;
                  break;
                }
              }
           	  if(z!=10)break;
          }
          if(y!=10)break;
      }
      if(j==0)printf("0");
      if(i!=n)printf("\n");
      j=0;
    }
	return 0;
}