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 <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
int main(){
    int i,j,k,l,x,y,z,n,count=0,f[10010]={};
    scanf("%d",&n);
    for(l=1; l<=n; l++){
        for(i=1; i<sqrt(1.0*l); i++){
            for(j=i; j<sqrt(1.0*l); j++){
                for(k=j; k<sqrt(1.0*l); k++){
                    if(i*i+j*j+k*k+i*j+j*k+k*i==l){
                        if(i!=j && j!=k && k!=i)f[l]+=6;
                        else if(i==j && j==k)f[l]++;
                        else{f[l]+=3;}
                    }
                }
            }
        }
    }
    for(i=1; i<=n; i++){
        printf("%d\n",f[i]);
    }
    return 0;
}