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::str::FromStr;

#[allow(dead_code)]
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 = read::<usize>();
    let mut candidate = vec![1];
    {
        let mut v9 = 1;
        for i in 0..20 {
            v9 *= 9;
            if v9 > n {
                break;
            }
            candidate.push(v9);
        }
        let mut v6 = 1;
        for j in 0..20 {
            v6 *= 6;
            if v6 > n {
                break;
            }
            candidate.push(v6);
        }
    }
    let mut ans = n;
    let mut dp = vec![1 << 20; n + 1];
    dp[0] = 0;
    for i in 0..n + 1 {
        for &v in candidate.iter() {
            let nv = v + i;
            if nv > n {
                continue;
            }
            dp[nv] = std::cmp::min(dp[nv], dp[i] + 1);
        }
    }
    println!("{}", dp[n]);
}