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: Runtime Error)

#include <stdio.h>
#include <string.h>

int a[10000][3];
int a_count = 0;

int main()
{
    int N;
    
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        printf("%d\n", findMatch(i));
    }
    
}

int findMatch(int target){
    memset(a, 0, sizeof(a));
    
    int count = 0, result = 0;
    int i, j, k;
    for (i = 1; (i * i) < target; i++) {
        for (j = 1; (j * j) < target; j++) {
            for (k = 1; (k * k) < target; k++) {
                //if (isEquiv(i, j, k)) return count * 3;
                result = calculate(i, j, k);
                if (result == target) {
                    count++;
                    a[a_count][0] = i;
                    a[a_count][1] = j;
                    a[a_count][2] = k;
                    a_count++;
                }
                if (result > target) break;
            }
        }
    }
    return count;
}

int calculate(int x, int y, int z){
    int n = x * x + y * y + z * z + x * y + y * z + z * x;
    return n;
}