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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include<algorithm>
#include<string>
#include <map>
#include <queue>
#include <stack>

#define DIV 1000000007
using namespace std;

int log_six(int x) {
    int ans = -1;
    while(x>0){
        ans++;
        x /= 6;
    }

    return ans;
}

int log_nine(int x) {
    int ans = -1;
    while (x > 0) {
        ans++;
        x /= 9;
    }

    return ans;
}
int main() {
    int N;
    cin >> N;

    vector<long long> cost(N + 1, LLONG_MAX);
    cost[0] = 0;
    int temp = 1;
    while(temp<= N) {
        cost[temp] = 1;
        temp *=6 ;       
    }
    temp = 1;
    while (temp <= N) {
        cost[temp] = 1;
        temp *= 9;
    }

    for (int i = 1; i <= N; i++) {
        int m_six = log_six(i);
        int m_nine = log_nine(i);
        //cout << "i=" << i << "  ";
        //cout << "m_six=" << m_six << "  m_nine=" << m_nine << "  ";
        cost[i] = min(cost[i],cost[i-pow(6,m_six)] +cost[pow(6,m_six)]);
        cost[i] = min(cost[i], cost[i - pow(9, m_nine)] + cost[pow(9, m_nine)]);
       // cout << cost[i] << endl;
    }
    cout << cost[N] << endl;
}