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;
	int na[10010];
	int count = 0;
	int fn=0;
	int i,j,k;
	
	scanf("%d",&n);
	
	for(i=0;i<=n;i++){
		na[i] = 0;
	}
	
	for(i=1;i<=n;i++){
		for(j=1;j<=n;j++){
			for(k=1;k<=n;k++){
				fn = (i*i)+(j*j)+(k*k)+(i*j)+(j*k)+(k*i);
				if(fn <= n){
					na[fn]=na[fn] + 1;
				}
			}
		}
	}
	
	for(i=1;i<=n;i++){
		printf("%d\n",na[i]);
	}
	
	return 0;
}