一瓶汽水1块钱，3个瓶盖换一瓶汽水，2个空瓶换一瓶汽水，假如手里有5块钱，能喝多少瓶汽水

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用三种方式实现以下命题：
一瓶汽水1块钱，3个瓶盖换一瓶汽水，2个空瓶换一瓶汽水，假如手里有5块钱，能喝多少瓶汽水

java
采用经典Java OOP方式实现，一目了然。

/**已喝掉*/
static int sodaDrinked = 0;
/**汽水*/
static int soda = 5;
/**瓶盖*/
static int cap = 0;
/**空瓶*/
static int emptyBottle = 0;


public static void main(String[] args){

    while (true) {
        if (soda > 0) {
            System.out.println("喝汽水");

            drink();
            pring();
            continue;
        }
        if (cap >= 3) {
            System.out.println("瓶盖换汽水");

            capToSoda();
            pring();
            continue;
        }
        if (emptyBottle >= 2) {
            System.out.println("空瓶换汽水");

            emptyBottleToSoda();
            pring();
            continue;
        }
        System.out.println("剩余: 瓶盖" + cap);
        System.out.println("剩余: 空瓶" + emptyBottle);
        System.out.println("喝掉：" + sodaDrinked);
        break;
    }

}

public static void drink() {
sodaDrinked++;
soda--;
cap++;
emptyBottle++;
}

public static void capToSoda() {
cap = cap - 3;
soda++;
}

public static void emptyBottleToSoda() {
    emptyBottle = emptyBottle - 2;
    soda++;
}

public static void pring() {
        System.out.println("喝掉" + sodaDrinked + " 瓶盖" + cap + " 空瓶" + emptyBottle + " 剩余汽水" + soda);
}

Excel
细心的QA妹子，使用Excel最终获得答案，大写的赞

Python
下面代码“脏乱差”，不建议初学者去尝试读，因为这要有一丢丢的Python语法基础。

    # -*- coding: utf-8 -*-
# @Time    : 18/7/26 上午11:56
# @Author  : Ligang.Wang[[email protected]]
# @File    : SodaHandle.py
#
    class SodaHandle():
        totalSoda = 0

        def change(self, soda, time, cap, bottle):
            if time == 0:
                cap = soda
                bottle = soda
                self.totalSoda += soda;
            else:
                cap += soda
                bottle += soda

            soda = 0
            soda += cap // 3
            cap =cap % 3
            soda += bottle // 2
            bottle = bottle % 2
            print(" The time", time, "you could exchange soda:", soda, "remainder bottle:", bottle, ",remainder caps:", cap)

            time += 1
            if (soda > 0):
                self.totalSoda += soda
                self.change(soda, time, cap, bottle)
            else:
                print(" Not enough caps and bottles to exchange!")

        def game(self):
            self.change(5, 0, 0, 0)
            print("You can drink ", self.totalSoda, "soda finally!")


    soda = SodaHandle()
    soda.game()

执行结果图：

java 迭代

    /**
     *@Author小浣熊
     */
     int p=5;
public void getInt(int k,int g){//k 是空瓶，g 指瓶盖,刚开始5块钱可以买5瓶，所以这里传5,5进来
    int aaa=k/2+k%2+g/3;//空
    int bbb=k/2+g/3+g%3;//盖
    p=p+k/2+g/3;
    if (aaa<2&&bbb<3){
        YUtils.i("12345","可以喝"+p+"瓶汽水");
    }else {
        getInt(aaa, bbb);
    }
}