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::collections::HashMap;
use std::io::*;
use std::str::FromStr;

//https://onlinejudge.u-aizu.ac.jp/solutions/problem/ALDS1_5_B/review/2709881/arsenic28/Rust
trait Joinable {
    fn join(self, sep: &str) -> String;
}

impl<U: ToString, T: Iterator<Item = U>> Joinable for T {
    fn join(self, sep: &str) -> String {
        self.map(|x| x.to_string()).collect::<Vec<_>>().join(sep)
    }
}

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: i64 = read();
    let n_sq = (n as f64).sqrt() as i64;
    let mut f_n = HashMap::new();

    for x in 1..=n_sq {
        for y in 1..=n_sq {
            for z in 1..=n_sq {
                let temp_n = x * x + y * y + z * z + x * y + x * z + y * z;
                //println!("x{} y{} z{} n{}", x, y, z, temp_n);
                let count = f_n.entry(temp_n).or_insert(0);
                *count += 1;
            }
        }
    }
    for i in 1..=n{
        println!("{:?}",f_n.get(&(i as i64)).unwrap_or(&0)); 
    }
}