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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int ruizyo(int x,int n){
	int i,ret;
	ret = 1;
	for(i=0;i<n;i++)
		ret = ret*x;
	return ret;
}

int main(){
	int count=0;
	long int num,temp,i;
	long int max_nine_number,max_six_number;
	scanf("%ld",&num);

	/* 累乗テーブル作成 */
	while(num>0){
		max_nine_number = 0;
		max_six_number = 0;

		/* 9の累乗のうち、超えない最大の数を探す*/
		for(i=1;i<num;i++){
			temp = ruizyo(9,i);
			if(temp<=num){
				max_nine_number = temp;
			}else{
				break;
			}	
		}
		/* 6の累乗のうち、超えない最大の数を探す*/
		for(i=1;i<num;i++){
			temp = ruizyo(6,i);
			if(temp<=num){
				max_six_number = temp;
			}else{
				break;
			}	
		}
		
		/* 1の累乗 */
		if(num < 6){
			num -= 1;
			count++;
		}
		/*12以上、15以下は6*/
		else if(num >= 12 && num <=15){
			num -=6;
			count++;
		}
		/* 6の累乗 */
		else if(max_six_number > max_nine_number){
			num -= max_six_number;
			count++;
		}
		/* 9の累乗 */
		else{
			num -= max_nine_number;
			count++;
		}
	}
	printf("%d\n",count);	
	return 0;
}