UVa 11150 Cola (water ver.)

本文介绍了一种通过借用额外空瓶来最大化享受饮料数量的方法。通过递归利用空瓶换取新饮料并再次产生的空瓶,可以实现比常规方法更多的饮料消费。

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11150 - Cola

Time limit: 3.000 seconds 

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2091

You see the following special offer by the convenience store:

" A bottle of Choco Cola for every 3 empty bottles returned "

Now you decide to buy some (say N) bottles of cola from the store. You would like to know how you can get the most cola from them.

The figure below shows the case where N = 8. Method 1 is the standard way: after finishing your 8 bottles of cola, you have 8 empty bottles. Take 6 of them and you get 2 new bottles of cola. Now after drinking them you have 4 empty bottles, so you take 3 of them to get yet another new cola. Finally, you have only 2 bottles in hand, so you cannot get new cola any more. Hence, you have enjoyed 8 + 2 + 1 = 11 bottles of cola.

You can actually do better! In Method 2, you first borrow an empty bottle from your friend (?! Or the storekeeper??), then you can enjoy 8 + 3 + 1 = 12 bottles of cola! Of course, you will have to return your remaining empty bottle back to your friend.

Input

Input consists of several lines, each containing an integer N (1 ≤ N ≤ 200).

Output

For each case, your program should output the maximum number of bottles of cola you can enjoy. You may borrow empty bottles from others, but if you do that, make sure that you have enough bottles afterwards to return to them.

Sample Input

8

Sample Output

12

Note: Drinking too much cola is bad for your health, so... don't try this at home!! :-) 


完整代码:

/*0.012s*/

#include<cstdio>

int main()
{
	int n, sum;
	while (~scanf("%d", &n))
	{
		sum = n;
		while (n > 2)
		{
			sum += n / 3;
			n -= n / 3 * 2;
		}
		if (n == 2) ++sum;
		printf("%d\n", sum);
	}
	return 0;
}

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