原始题目

C - Strange Bank

• Time limit : 2sec
• Memory limit : 256MB
• 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, 62(=36) yen, 63(=216) yen, ...

9 yen, 92(=81) yen, 93(=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≤N≤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(=62) yen and 81(=92) 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

题目大意

规定三种取钱的方法，一次可以取1元，6^k元或者9k元 （k>=1） ，问取n元最少的次数是多少。

解题思路

将n拆成i和n-i。分别用6和9进制表示，维护系数和最小值即可。

解题代码

#include <cstdio>
#include <iostream> 
using namespace std;

int n;
int main()
{
	while(~scanf("%d" ,&n))
	{
		int res=n; //最大情况肯定为n次  全部每次取1元 
		for(int i=0;i<=n;i++)
		{
			int cc=0;
			int t=i;
			while(t) cc+=t%6,t/=6;  //计算6进制的系数和 
			t=n-i;
			while(t) cc+=t%9,t/=9;  //计算9进制的系数和 
			if(res>cc) res=cc; 
		
		}
		printf("%d\n",res);
	}

}

收获与反思

• 理解题意，三种方法算最小值其实就是求一个数用6/9进制混合表达的系数最小值。

• 把n拆成两部分，维护系数最小值。

  while(t) cc+=t%k,t/=k;

• 该方法得到的cc为t用k进制表达的系数和。