当前位置: 首页 > ops >正文

《Python语言程序设计》第4章第7题,这次利用之前学过的第7章的内容使用对象和类,来修改这道题

在这里插入图片描述


original codeamount = eval(input('Enter an amount, for example,11.56: '))#偷懒直接做了输入
amount = 11.56remain_in_amount = int(amount * 100)number_of_one_dollars = remain_in_amount // 100
remain_in_amount = remain_in_amount % 100number_of_quarters = remain_in_amount // 25
remain_in_amount = remain_in_amount % 25number_of_dimes = remain_in_amount // 10
remain_in_amount = remain_in_amount % 10number_of_nickels = remain_in_amount // 5
remain_in_amount = remain_in_amount % 5number_of_pennies = remain_in_amountif amount >1:amount_text= 'consists'
elif amount==1:amount_text= 'consist'print('Your amount',amount,'consists of\n','\t', number_of_one_dollars,'dollars\n','\t',number_of_quarters,'quarters\n','\t',number_of_dimes,'dimes\n','\t',number_of_nickels,'nickels\n','\t',number_of_pennies,'pennies')

原始的输出结果

在这里插入图片描述

我们使用第七章对象和类的功能

请用户输入钱数
将输入的整数或者浮点数进行整数化

class amount_com:def __init__(self, amount):self.amount = amountdef get_int_amount(self):text_nickels = ''text_dollars = ''text_quarters = ''text_dimes = ''text_pennies = ''remain_in_amount = int(self.amount * 100)number_of_one_dollars = remain_in_amount // 100remain_in_amount = remain_in_amount % 100number_of_quarters = remain_in_amount // 25remain_in_amount = remain_in_amount % 25number_of_dimes = remain_in_amount // 10remain_in_amount = remain_in_amount % 10number_of_nickels = remain_in_amount // 5remain_in_amount = remain_in_amount % 5number_of_pennies = remain_in_amountif number_of_one_dollars == 1:text_dollars = 'dollar'elif number_of_one_dollars > 1:text_dollars = 'dollars'if number_of_quarters == 1:text_quarters = 'quarter'elif number_of_quarters > 1:text_quarters = 'quarters'if number_of_nickels == 1:text_nickels = 'nickel'elif number_of_nickels > 1:text_nickels = 'nickels'if number_of_dimes == 1:text_dimes = 'dime'elif number_of_dimes > 1:text_dimes = 'dimes'if number_of_pennies == 1:text_pennies = 'pennie'elif number_of_pennies > 1:text_pennies = 'pennies'print(f"""Your amount {self.amount} consists of:{number_of_one_dollars}, {text_dollars}{number_of_quarters}, {text_quarters}{number_of_dimes}, {text_dimes}{number_of_nickels}, {text_nickels}{number_of_pennies}, {text_pennies}""")num = 11.56
a = amount_com(num)
a.get_int_amount()

在这里插入图片描述
运行效果。

http://www.xdnf.cn/news/8442.html

相关文章:

  • 仿真APP助力提升卡车驾驶室驾乘舒适度与安全性
  • 模型压缩,AWQ与GPTQ量化方法分析
  • 【信息系统项目管理师】第20章:高级项目管理 - 28个经典题目及详解
  • 学习日记-day14-5.23
  • Redis淘汰策略
  • vue pinia 独立维护,仓库统一导出
  • 虚拟机Centos7:Cannot find a valid baseurl for repo: base/7/x86_64问题解决
  • Linux 下使用 Sysbench 进行性能测试
  • ConceptAttention:Diffusion Transformers learn highly interpretable features
  • Tailwind css实战,基于Kooboo构建AI对话框页面(一)
  • 纸牌游戏(基于集合,和自定义排序实现)
  • linux_cmake的笔记
  • 从底层原理分析Python 常用字符串拼接方法效率差异
  • BeeWorks局域网聊天工具:打造智能高效的企业级即时通讯新生态
  • Spring生态的核心思想
  • 每周资讯 | 网易游戏发布会9款新品曝光;TikTok小程序测试开启
  • LangChain4j入门AI(八)使用RAG增加私有知识库
  • 初识 Flask 框架
  • Sa-Token登录权限认证
  • 水力发电站人员定位新方案
  • 企业为什么要实施信息化体系建设
  • CDGP|数据治理:如何最大化企业数据中台商业价值?
  • 如何创建WooCommerce子主题
  • 九州未来十三载:开源赋能 智启未来
  • Qt 的多线程
  • 记录一次请求数据很慢的灾难
  • 每日算法 -【Swift 算法】反转整数的陷阱与解法:Swift 中的 32 位整数处理技巧
  • 列表集合字典推导式
  • MOSFET雪崩击穿
  • 似然分布与共轭分布,算是补作业吧