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>
int main () {
    int N;
    scanf("%d",&N);
    int b[N];
    int i,x,y,z,m;
    for(i=0;i<N;i++){
        b[i] = 0;
        for(x=1;x<=100;x++){
            for(y=x;y<=100;y++){
                for(z=y;z<=100;z++){
                    m = x*x + y*y + z*z + x*y + y*z + z*x;
                    if(m == (i+1)){
                        if(x == y && y == z){
                            b[i] += 1;
                        }else if(x != y && y != z && z != x){
                            b[i] += 6;
                        }else{
                            b[i] += 3;
                        }
                    }
                }
            }
        }
        printf("%d\n",b[i]);
    }
    return 0;
}