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: Accepted)

#include<stdio.h>
#include<stdlib.h>

int main(void)
{
    int N;
    if(scanf("%d", &N) == EOF) exit(1);

    int i, x, y, z, count = 0;
    for (i = 1; i <= N; i++){
        for (x = 1; x < 100 ;  x++){
            if(x*x + 2*x + 3 > i) continue;

            for (y = x ; y < 100 ; y++){
                if(y*y + x*x + x*y + y + x + 1 > i) continue;

                for (z = y; z < 100 ; z++){
                    if (x*x+y*y+z*z+x*y+y*z+z*x > i) {
                        continue;
                    }else if (x*x+y*y+z*z+x*y+y*z+z*x == i) {
                        count++;

                        if ((x != y) && (y != z) && (z !=x)){
                            count = count + 5;
                        } else if ((x != y) || (y != z) || (z !=x)){
                            count = count + 2;
                        }
                    }
                }
            }
        }
        printf("%d\n", count);
        count =0;
    }

    return 0;
}