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: Accepted)

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

int main(void){
	int N, i,v,n;
  int x, y, z;
   int ans[10000];
  
  
  scanf("%d", &N);
   n=sqrt(N);
  for(i = 0;i<=N;i++){
    ans[i] = 0;
  }  

   for(x = 1; x <=n;x++){
    for(y= 1; y<=n; y++){
      	for(z = 1; z<=n; z++){
           v = (x*x)+(y*y)+(z*z)+(x*y)+(y*z)+(z*x);
        	if(v<=N){
          		ans[v] = ans[v]+1;
            }
      	}
    }
   }
    for(i = 1; i<=N; i++){
    printf("%d\n", ans[i]);
  }
    return 0;
}