# Cola
題目連結 [UVA 10050](https://onlinejudge.org/external/111/11150.pdf)
## 中文簡述

測資會給予多組可樂瓶數，每三個空瓶可兌換一瓶可樂，計算最多可享用多少可樂。（可與朋友借空瓶來兌換可樂，但最後要能夠歸還空瓶）

## [think]
1. 用迴圈計算分別借 0~2 個空瓶時能享用的可樂數

2. 檢查最後是否能歸還空瓶，若否，則此種兌換方法不成立

## solution:

#include <iostream>
using namespace std;

int main()
{
    int num;
    while (cin >> num)
    {
        int max= 0;
        
        for (int i = 0; i < 3; i++)
        { 
            // number of empty bottle I borrow 
            int drink=num; 
            int tmp_cola = 0;
            int tmp_empty = cola_num + i; 

            while (tmp_empty >= 3)
            { 
                drink_cola_sum += tmp_empty / 3;
                tmp_empty = tmp_empty % 3 + tmp_empty / 3;
            }

            // can't return empty bottle to friend => can't borrow i bottles
            if (tmp_empty < i)
            {
                drink_cola_sum = 0;
            }

            if (max_cola < drink_cola_sum)
            {
                max_cola = drink_cola_sum;
            }
        }
        cout << max_cola << endl;
    }

    return 0;
}

###### tags: `UVA`
回目錄 [學習筆記](/gIBZqAbWTCis7uOPp149gA)