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)

//AC
#include <stdio.h>


int main(void){

    int N;
    scanf("%d",&N);

    for(int n=1;n<=N;n++){ 
        int count=0;

        for(int i=1;i*i+2*i<=n;i++){
            for(int j=1;j*j+2*j<=n&&j<=i;j++){
                for(int k=1; k*k+2*k<=n&&k<=j;k++){
                    if(i*i+j*j+k*k+i*j+j*k+k*i==n){
                        if(i==j&&i==k){count++;}
                        else if(i!=j&&j!=k&&k!=i){count+=6;}
                        else{count+=3;}
                }
            }
        }
        printf("%d\n",count);
    }
        
    }

}