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(void){
    int n;
    scanf("%d",&n);
    register int i,j,k,l;
    int ans[n];
    for(i = 0;i < n;i++)ans[i] = 0;
    for(i = 3;i <= 102;i++){
        for(j = 1;j <= i-2;j++)for(k = 1;k <= i-1-j;k++){
            l = i - j - k;
            int tn = i * i - j * k - k * l - j * l;
            if(tn > 0 && tn <= n)ans[tn-1]++;
        }
    }
    for(i = 0;i < n;i++)printf("%d\n",ans[i]);
}