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>
#include <math.h>

int main(void){
    int N;
    int tmp;
    int tmp2;
    int ni;
    int i,j,k;
    int count =0;
    int f[10002];

    scanf("%d", &N);
    for (ni=1; ni<=N; ni++) {
        //tmp = (int)sqrt((double)ni/6.0)+10;
        tmp = 90;
        //tmp = (int)sqrt(ni-2);
        for (i=1; i<=tmp; i++) {
            for (j=1; j<=tmp; j++) {
                for (k=1; k<=tmp; k++) {
                    tmp2 = i*i + j*j + k*k + i*j + j*k + k*i;
                    if (tmp2==ni){
                        count = count +1;
                    }
                }
            }
        }
        f[ni] = count;
        count = 0;


    }

    for (i=1; i<=N; i++) {
        printf("%d\n", f[i]);
    }

    

    return 0;
}