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 = 0;
    printf("%d",&N);
    int answer[100];
        int x = 1, y = 1, z = 1;
        int n = 1;
        int a = 0;
        int count = 0;
        for (int i = N; n <= i; i++)
        {
            count = 0;
            while (x <= 100)
            {
                while (y <= 100)
                {
                    while (z <= 100)
                    {
                        if (x * x + y * y + z * z + x * y + y * z + z * x == n)
                        {
                            count++;
                        }
                        z++;
                    }
                    y++;
                }

                x++;
            }
            answer[a] = count;
            x = 1;
            y = 1;
            z = 1;
        }
        for(int i = 0;i <= N;i++)
        {
            printf("%d\n",answer[i]);
        }
        
    return 0;
}