UVA11150 Cola【数学】

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.

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

Sample Input

Sample Output

问题链接：UVA11150 Cola

问题简述：（略）

问题分析：

喝可乐，每3个空瓶可以换一瓶可乐，迭代计算问题。

如果最后剩下2瓶则可以换一瓶。喝掉这2瓶再从朋友那里借一个空瓶，3个空瓶换一瓶，喝了之后，将空瓶还给朋友。

既然2瓶可以喝3瓶，那直接用数学公式算就可以了，时间复杂度为O(1)。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA11150 Cola */#include <bits/stdc++.h>using namespace std;int main()
{int n;while(~scanf("%d", &n))printf("%d\n", (n / 2 )* 3 + n % 2);return 0;
}

AC的C++语言程序（模拟法）如下：

/* UVA11150 Cola */#include <bits/stdc++.h>using namespace std;int main()
{int n, ans;while(~scanf("%d", &n)) {ans = n;while(n >= 3) {ans += n / 3;n = n / 3 + n % 3;}if(n == 2)ans++;printf("%d\n", ans);}return 0;
}

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