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(){
    int x,y,z,n,i;
    scanf("%d",&n);
    if(n<=5){
        for(i=0;i<n;i++)
        printf("0\n");
    }else{
    for(i=1;i<=n;i++){
    for(z=1;z<n;z++){
        for(y=1;y<n;y++){
            for(x=1;x<n;x++){
                if((x+y+z)^2-(x*y+y*z+z*x)==n){
                    break;
                }
            }
        }
    }
    if((x+y+z)^2-(x*y+y*z+z*x)==n&&(x==y||y==z)){
        if(x==y&&y==z) printf("1\n");
        else printf("3\n");
    }else if((x+y+z)^2-(x*y+y*z+z*x)==n&&(x!=y&&y!=z&&z!=x)){
        printf("6\n");
    }else
        printf("0\n");
    }
    }
    return 0;
}