10元最多可喝多少瓶啤酒？（不可借酒+可借酒）

背景

《爱情公寓5》中有个剧情：每瓶啤酒2元，2个空酒瓶或4个瓶盖可换1瓶啤酒。10元最多可喝多少瓶啤酒？
脑海模拟起来的确有点费劲。心算结果是15瓶，而剧情实践竟然是20瓶！结合弹幕说的酒吧可能能够借酒，故猜想借酒使最终喝了20瓶。
若是改变拥有的钱数或啤酒价格时，答案又是什么呢？
此时，不由想用编程的方法解决一下。

思路

分两种状况：酒吧能够借酒；酒吧不能够借酒；

酒吧不能够借酒

此时，当剩余酒盖数<4且剩余空瓶数<2时，计算就结束了

酒吧能够借酒

这种状况比较麻烦，须要考虑1个空瓶和3个瓶盖时借一瓶酒的状况。此时，当换完酒、假设喝了酒并把酒瓶换新酒、喝了新酒以后（空瓶1，瓶盖3），手上的酒瓶和酒盖的价值不大于已借的酒瓶数（2瓶）时，就须要考虑结束计算了。
详细代码以下：

//
//  main.swift
//  beerMaxDrink
//
//  Created  on 2020/2/4.
//  Copyright © 2020. All rights reserved.
//
/*
 酒吧啤酒2元一瓶，两个空瓶或四个瓶盖能够换一瓶啤酒（酒吧概不借酒），你有10元钱，请问最多能够喝几瓶？
 */

import Foundation

//共累计喝的瓶数
var drinkSum = 0
//当前瓶子数
var bottleNum = 0
//当前瓶盖数
var capsNum = 0

//返回（累计喝瓶数，剩余瓶子数，剩余瓶盖数）
func getMaxDrinkSum(money:Float, price:Float) ->(Int, Int, Int) {

    bottleNum = Int(money / price)
    capsNum = bottleNum
    drinkSum = bottleNum
    
    //开始换酒,是个循环
    while bottleNum > 1 || capsNum > 3 {
        //酒瓶换
        let wineAddedByBottle = bottleNum / 2
        drinkSum += wineAddedByBottle
        bottleNum = bottleNum % 2 + wineAddedByBottle
        capsNum += wineAddedByBottle
        
        //酒盖换
        let wineAddedByCaps = capsNum / 4
        drinkSum += wineAddedByCaps
        capsNum = capsNum % 4 + wineAddedByCaps
        bottleNum += wineAddedByCaps
    }
    return (drinkSum, bottleNum, capsNum)
}

//能够借酒时，返回（累计喝瓶数，剩余瓶子数，剩余瓶盖数）
func getMaxDrinkSumCanBorrow(money:Float, price:Float) ->(Int, Int, Int) {

    var borrowedNum = 0
    bottleNum = Int(money / price)
    capsNum = bottleNum
    drinkSum = bottleNum
    
    //开始换酒,是个循环
    while bottleNum >= 1 || capsNum >= 3 {
 
        //酒瓶换
        let wineAddedByBottle = bottleNum / 2
        drinkSum += wineAddedByBottle
        bottleNum = bottleNum % 2 + wineAddedByBottle
        capsNum += wineAddedByBottle
        
        //若是（2酒瓶2酒盖时，"酒瓶换"已经把2酒瓶换为1瓶1盖，即变为1瓶3盖,方便作判断）借一瓶后，空瓶和瓶盖能换的酒数 <= 已借瓶数时，结束;不然，借一瓶
        if 1 == bottleNum || 3 == capsNum{//不能少，不然在循环时会提早借酒
            if (bottleNum + 1)/2 + (capsNum + 1)/4 <= borrowedNum {
                
                //不该该喝一瓶，回退
                bottleNum += 1
                capsNum -= 1
                return (drinkSum - 1, bottleNum - borrowedNum, capsNum - borrowedNum)
            }else{ //借一瓶
                borrowedNum += 1
                bottleNum += 1
                capsNum += 1
                
                //酒瓶换
                let wineAddedByBottle = bottleNum / 2
                drinkSum += wineAddedByBottle
                bottleNum = bottleNum % 2 + wineAddedByBottle
                capsNum += wineAddedByBottle
            }
        }
        
        //酒盖换
        let wineAddedByCaps = capsNum / 4
        drinkSum += wineAddedByCaps
        capsNum = capsNum % 4 + wineAddedByCaps
        bottleNum += wineAddedByCaps
    }
    return (drinkSum, bottleNum, capsNum)
}

let rslt0 = getMaxDrinkSum(money: 10.0, price: 2)
print(rslt0)

let rsltCanBorrow0 = getMaxDrinkSumCanBorrow(money: 10.0, price: 2)
print(rsltCanBorrow0)

let rslt1 = getMaxDrinkSum(money: 10.0, price: 5)
print(rslt1)

let rsltCanBorrow1 = getMaxDrinkSumCanBorrow(money: 10.0, price: 5)
print(rsltCanBorrow1)

运行结果以下，证实代码正确：

(15, 1, 3)
(20, 0, 0)
(3, 1, 3)
(8, 0, 0)
Program ended with exit code: 0