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：
1 #include<stdio.h> 2 int main() 3 { 4 int n; 5 while (scanf("%d", &n) == 1 && n) 6 { 7 int a = 0, b = 0; 8 int count = 0; 9 if (n == 1 || n == 2) //两种特殊情况 10 { 11 count = n - 1; 12 } 13 while ((n + b) >= 3) 14 { 15 a = n + b; 16 n = (n + b) / 3; //n+b为目前的所有空瓶子数 17 b = a % 3; 18 count += n; //目前换取到的可乐数 19 20 if (n + b == 2) //如果最后还剩下两个空瓶子就可以先借一 21 //个空瓶子 22 count += 1; //从而又可以多得一瓶可乐 23 } 24 25 26 printf("%d\n", count); 27 } 28 29 30 return 0; 31 }
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！
