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 Rust (1.42.0) (judge verdict: Accepted)

use std::cmp;
use std::fmt::Debug;
use std::str::FromStr;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where
    T: FromStr,
    <T as FromStr>::Err: Debug,
{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim()
        .split_whitespace()
        .map(|c| T::from_str(c).unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_as_string() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn calc(x: usize, y: usize, z: usize) -> usize {
    return x * x + y * y + z * z + x * y + y * z + z * x;
}

fn main() {
    let n: usize = read_as_vec()[0];

    let mut v: Vec<usize> = vec![0; n + 1];

    for x in 1..=100 {
        for y in 1..=100 {
            for z in 1..=100 {
                let f = calc(x, y, z);
                if f < v.len() {
                    v[f] += 1;
                }
            }
        }
    }
    for i in 1..v.len() {
        println!("{}", v[i]);
    }

    // let mut t: Vec<Vec<(usize, usize, usize)>> = vec![vec![]; n + 1];
    // v[6] = 1;
    // t[6].push((1, 1, 1));
    // for i in 6..=n {
    //     for ti in 0..t[i].len() {
    //         let tr = t[i][ti];
    //         if tr.0 == tr.1 && tr.1 == tr.2 {
    //             let j = calc(tr.0 + 1, tr.1, tr.2);
    //             v[j] += 3;
    //             t[j].push((tr.0 + 1, tr.1, tr.2));
    //         } else {
    //         }
    //     }
    // }
}