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
8
Sample Output
12
其实，这个问题。最上面的可乐，如果刚喝两瓶就拿来换是最聪明的选择。两瓶换三瓶。
意思就是，喝一瓶，count++，空瓶++，空瓶到两瓶，就马上去换。
#include<iostream> using namespace std; int count = 0; int main() { int n; while(cin>>n){ int m = 0; count = 0; while(n > 0 || m >= 2){ if(n){ // 可乐少了一瓶 n--; //喝了一瓶 count++; //空瓶多了一瓶 m++; } if(m >= 2){ //空瓶两瓶，借一瓶换一瓶，还了，相当于0 m=0; //又喝了刚才换得一瓶 count++; } } cout<<count<<endl; } return 0; }然后今早醒来，突然想到，不就是两个空瓶换一个？??有那么麻烦么！！！！
#include<iostream> using namespace std; int main() { int n; while(scanf("%d",&n)!= EOF){ cout<<n+n/2<<endl; } return 0; }其实还可以这样
#include<iostream> using namespace std; int main() { int n; while(scanf("%d",&n)!= EOF){ cout<<(n/2)*3+(n%2)<<endl; } return 0; }
