Mbp通过筛选器和中间件实现异常,日志,事务及接口返回数据格式化aop处理.
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,而剧情实践居然是20瓶!结合弹幕说的酒吧可能可以借酒,故猜测借酒使最终喝了20瓶。
如果改变拥有的钱数或啤酒价格时,答案又是什么呢?
此时,不禁想用编程的方法解决一下。
思路
分两种情况:酒吧可以借酒;酒吧不可以借酒;
CTF–HTTP服务–SSI注入
酒吧不可以借酒
此时,当剩余酒盖数<4且剩余空瓶数<2时,计算就结束了
酒吧可以借酒
这种情况比较麻烦,需要考虑1个空瓶和3个瓶盖时借一瓶酒的情况。此时,当换完酒、假设喝了酒并把酒瓶换新酒、喝了新酒之后(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,就需要考虑结束计算了。
详细代码如下:
//
// main.swift
// beerMaxDrink
//
// Created by 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可以换一瓶啤酒(酒吧概不借酒),你有10元钱,请问最多可以喝几瓶?
*/
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)
}
//可以借酒时,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
//如果(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)借一瓶后,空瓶和瓶盖能换的酒数 <= 已借瓶数时,结束;否则,借一瓶
if 1 == bottleNum || 3 == capsNum{//不能少,否则在循环时会提前借酒
if (bottleNum + 1)/2 + (capsNum + 1)/4 <= borrowedNum {
//不应该喝一瓶,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)
}
for i in 0..<20{
let rslt0 = getMaxDrinkSum(money: Float(i), price: 2)
print("-----------------------")
print("\(i)元:不能借酒:\(rslt0)")
let rsltCanBorrow0 = getMaxDrinkSumCanBorrow(money: Float(i), price: 2)
print("能借酒:\(rsltCanBorrow0)")
}
let rslt1 = getMaxDrinkSum(money: 10.0, price: 5)
print("-----------------------")
print("酒价格5元时,不能借酒:\(rslt1)")
let rsltCanBorrow1 = getMaxDrinkSumCanBorrow(money: 10.0, price: 5)
print("能借酒:\(rsltCanBorrow1)")
运行结果如下,证明代码正确。且可以看出规律:可以借酒时,一般比不能借酒时,可以多喝5瓶酒:
-----------------------
0元:不能借酒:(0, 0, 0)
能借酒:(0, 0, 0)
-----------------------
1元:不能借酒:(0, 0, 0)
能借酒:(0, 0, 0)
-----------------------
2元:不能借酒:(1, 1, 1)
能借酒:(4, 0, 0)
-----------------------
3元:不能借酒:(1, 1, 1)
能借酒:(4, 0, 0)
-----------------------
4元:不能借酒:(3, 1, 3)
能借酒:(8, 0, 0)
-----------------------
5元:不能借酒:(3, 1, 3)
能借酒:(8, 0, 0)
-----------------------
6元:不能借酒:(7, 1, 3)
能借酒:(12, 0, 0)
-----------------------
7元:不能借酒:(7, 1, 3)
能借酒:(12, 0, 0)
-----------------------
8元:不能借酒:(11, 1, 3)
能借酒:(16, 0, 0)
-----------------------
9元:不能借酒:(11, 1, 3)
能借酒:(16, 0, 0)
-----------------------
10元:不能借酒:(15, 1, 3)
能借酒:(20, 0, 0)
-----------------------
11元:不能借酒:(15, 1, 3)
能借酒:(20, 0, 0)
-----------------------
12元:不能借酒:(19, 1, 3)
能借酒:(24, 0, 0)
-----------------------
13元:不能借酒:(19, 1, 3)
能借酒:(24, 0, 0)
-----------------------
14元:不能借酒:(23, 1, 3)
能借酒:(28, 0, 0)
-----------------------
15元:不能借酒:(23, 1, 3)
能借酒:(28, 0, 0)
-----------------------
16元:不能借酒:(27, 1, 3)
能借酒:(32, 0, 0)
-----------------------
17元:不能借酒:(27, 1, 3)
能借酒:(32, 0, 0)
-----------------------
18元:不能借酒:(31, 1, 3)
能借酒:(36, 0, 0)
-----------------------
19元:不能借酒:(31, 1, 3)
能借酒:(36, 0, 0)
-----------------------
酒价格5元时,不能借酒:(3, 1, 3)
能借酒:(8, 0, 0)
Program ended with exit code: 0
TypeScript——基本类型
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