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>

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

  
  
  for(i = 1; i<N+1; i++){
      for(x = 1; x <(N^(1/2));x++){
    	for(y= 1; y< (N^(1/2)); y++){
      		for(z=1; z< (N^(1/2)); z++){
              	v = (x*x)+(y*y)+(z*z)+(z*y)+(y*x)+(x*z);
        		if(v==i && v<10000)
          			ans[i-1] = ans[i-1]+1;
      		}
    	}
  	}	
    printf("%d\n", ans[i-1]);
  }
    return 0;

}