Problem:
Score : 300 points

Problem Statement

To make it difficult to withdraw money, a certain bank allows its customers to withdraw only one of the following amounts in one operation:

1 yen (the currency of Japan)

6 yen, 6^2(=36) yen, 6^3(=216) yen, ...

9 yen, 9^2(=81) yen, 9^3(=729) yen, ...

At least how many operations are required to withdraw exactly N yen in total?

It is not allowed to re-deposit the money you withdrew.

Constraints

1 \leq N \leq 100000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If at least x operations are required to withdraw exactly N yen in total, print x.

Sample Input 1

127

Sample Output 1

4

By withdrawing 1 yen, 9 yen, 36(=6^2) yen and 81(=9^2) yen, we can withdraw 127 yen in four operations.

Sample Input 2

3

Sample Output 2

3

By withdrawing 1 yen three times, we can withdraw 3 yen in three operations.

Sample Input 3

44852

Sample Output 3

16

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::cmp::min;

macro_rules! parse_line {
    ($($t: ty),+) => ({
        let mut a_str = String::new();
        io::stdin().read_line(&mut a_str).expect("read error");
        let mut a_iter = a_str.split_whitespace();
        (
            $(
                a_iter.next().unwrap().parse::<$t>().expect("parse error"),
            )+
        )
    })
}

fn main() {
    let (n,) = parse_line!(u32);
    let mut dp = Vec::new();
    dp.resize((n as usize) + 1, n);
    for i in 0..min(n as usize, 3) {
        dp[i] = i as u32;
    }

    let max_log6 = (n as f32).log(6.0).floor() as usize + 1;
    let max_log9 = (n as f32).log(9.0).floor() as usize + 1;
    for i in 3..((n + 1) as usize) {
        let mut t = 1;
        for _ in 1..max_log6 {
            t *= 6;
            if i < t {
                break
            }
            dp[i] = min(dp[i], dp[i - t] + 1);
        }

        t = 1;
        for _ in 1..max_log9 {
            t *= 9;
            if i < t {
                break
            }
            dp[i] = min(dp[i], dp[i - t] + 1);
        }
    }

    println!("{}", dp[n as usize]);
}