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)

#![allow(unused)]

//input
//use std::io::Read;

//collections
//use std::collections::BTreeMap;
use std::collections::BTreeSet;
//use std::collections::HashMap;
//use std::collections::HashSet;

//入力マクロここから
//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
//入力マクロここまで

const INI: u64 = 1_000_000;

use std::cmp::*;

fn main() {
    input! {
        n: usize,
    }

    let mut y: BTreeSet<usize> = BTreeSet::new();

    for i in 0.. {
        if 6_usize.pow(i) > 100_000 {
            break;
        }
        y.insert(6_usize.pow(i));
    }

    for i in 0.. {
        if 9_usize.pow(i) > 100_000 {
            break;
        }
        y.insert(9_usize.pow(i));
    }

    //let y: Vec<u64> = y.iter().collect::<Vec<u64>>();
    let mut unit = Vec::new();

    for e in y {
        unit.push(e);
    }

    let mut dp = vec![INI; n + 1];

    dp[0] = 0;
    dp[1] = 1;

    for i in 0..n + 1 {
        for j in unit.iter() {
            if i + j > n {
                break;
            } else {
                dp[i + j] = min(dp[i + j], dp[i] + 1);
            }
        }
    }

    println!("{}", dp.last().unwrap());
}