空瓶子换可乐

 题目：

      Once upon a time, there is a special coco-cola store. If you return three empty bottles to the shop,
you’ll get a full bottle of coco-cola to drink. If you have n empty bottles right in your hand, how many
full bottles of coco-cola can you drink?

Input
There will be at most 10 test cases, each containing a single line with an integer n (1 ≤ n ≤ 100). The
input terminates with n = 0, which should not be processed.

Output
For each test case, print the number of full bottles of coco-cola that you can drink.
Spoiler
Let me tell you how to drink 5 full bottles with 10 empty bottles: get 3 full bottles with 9 empty
bottles, drink them to get 3 empty bottles, and again get a full bottle from them. Now you have 2
empty bottles. Borrow another empty bottle from the shop, then get another full bottle. Drink it, and
finally return this empty bottle to the shop!

Sample Input
3
10
81
0
Sample Output
1
5
40

思路分析：

     使用的方法是取除数跟取余数。每次循环用当前的总空瓶子n来除以3就可以得到当前能换取的可乐瓶数，然后记录此时的余数b，那么当前的总空瓶数就是n+b。特殊的情况要考虑，就是还剩两个空瓶子时可以先借一个空瓶子，这样就多得到一个可乐。

源代码：

#include<stdio.h>
int main()
{
    int n;
    while (scanf("%d", &n) == 1 && n)
    {
        int a = 0, b = 0;
        int count = 0;
        if (n == 1 || n == 2)          //两种特殊情况
        {
            count = n - 1;                  
        }
        while ((n + b) >= 3)
        {
            a = n + b;
            n = (n + b) / 3;          //n+b为目前的所有空瓶子数
            b = a % 3;
            count += n;              //目前换取到的可乐数

            if (n + b == 2)          //如果最后还剩下两个空瓶子就可以先借一
                                                           //个空瓶子
                count += 1;          //从而又可以多得一瓶可乐
        }


        printf("%d
", count);
    }


    return 0;
}

心得：

         这次比赛心态很不好，就这道题本来是简单的题，却很久才做出来，当时心里已经很浮躁了，后面的题更是无心去看了，结果整场比赛就做出这一道题，╮(╯▽╰)╭简直不能用文字形容自己的心情，经历了才知道。虽然自己不是那么在乎排名，但是浮躁的心态真的是个很大的障碍。不放弃！干吧得！