days[:int(a[:2]) - 1]为什么[后有:

时间: 2024-01-06 14:06:06 浏览: 18
这里的 `:` 表示切片操作。切片操作可以用来获取序列(如列表、字符串等)的一个子序列。语法为 `[start:end:step]`,其中 `start` 表示切片的起始位置,`end` 表示切片的结束位置(但不包括该位置对应的元素),`step` 表示切片的步长。 在这个代码中,`days` 是一个列表,`a` 是一个字符串,`a[:2]` 表示取 `a` 字符串的前两个字符,即日期的天数。这个值减去 1 后,就是要切片的结束位置。因此,`days[:int(a[:2]) - 1]` 表示取 `days` 列表的前面若干个元素,以构建一个新的列表。由于起始位置没有指定,因此默认为 0。而 `:` 之前也未指定结束位置和步长,因此默认为列表的末尾和步长为 1。
相关问题

猴子吃桃程序_扩展2 猴子第一天摘了若干个桃子,吃了一半,不过瘾,又多吃了1个。第二天早上将剩余的桃子又吃掉一半,并且又多吃了1个。此后每天都是吃掉前一天剩下的一半零一个。到第n天再想吃时,发现只剩下1个桃子,问第一天它摘了多少桃子?为了加强交互性,由用户输入不同的天数n进行递推,即假设第n天的桃子数为1。同时还要增加对用户输入数据的合法性验证(如:不允许输入的天数是0和负数) 程序运行结果示例: Input days: 0↙ Input days: -5↙ Input days: a↙ Input days: 3↙ x=10 输入格式:"%d" 输出格式: 输入提示信息:"Input days:\n" 输出:"x=%d\n"

以下是符合要求的程序代码: ```c #include <stdio.h> int main() { int n, x = 1; printf("Input days:\n"); while (scanf("%d", &n) != 1 || n <= 0) { // 输入合法性验证 printf("Input days:\n"); while (getchar() != '\n') continue; // 清空缓冲区 } for (int i = 1; i < n; i++) { x = (x + 1) * 2; } printf("x=%d\n", x); return 0; } ``` 程序的运行流程如下: 1. 程序首先输出提示信息 `"Input days:\n"`,并且使用 `scanf` 函数读入用户输入的天数 `n`,同时进行输入合法性验证,只有在用户输入的天数合法时才继续执行下面的代码。 2. 程序使用循环从第一天开始递推到第 `n-1` 天,每天的桃子数目为 `(x+1)*2`,其中 `x` 表示前一天剩余的桃子数。 3. 循环结束后,程序输出 `x` 的值,即第一天摘的桃子数目。 4. 程序返回0,表示正常结束。 需要注意的是,本程序还增加了输入合法性验证的功能,即只有当用户输入的天数为正整数时,才进行计算。如果用户输入的天数不合法,程序会输出提示信息 `"Input days:\n"`,并要求用户重新输入。同时,为了避免输入缓冲区中可能还有其他非法字符,程序在每次输入不合法时都会清空输入缓冲区。

invalid literal for int() with base 10: '45 days'

This error occurs because you are trying to convert the string '45 days' into an integer using the int() function, but '45 days' is not a valid integer. If you want to extract the number '45' from the string '45 days', you can use string manipulation techniques to extract the numeric part of the string and convert it to an integer. For example, you could use the split() method to split the string into a list of words, and then use indexing to extract the first word (which should be the numeric part of the string). Then you can use int() to convert that string to an integer: ``` my_string = '45 days' my_number = int(my_string.split()[0]) ``` This will set `my_number` to the integer value 45. Alternatively, if you only want to extract the numeric part of the string and don't need to convert it to an integer, you can use regular expressions to match the numeric part of the string: ``` import re my_string = '45 days' match = re.search(r'\d+', my_string) if match: my_number = match.group() ``` This will set `my_number` to the string value '45'.

相关推荐

Write a C++ program that defines a class DateV3 with the following: (1) private member variables: int year, month, day; (2) Has three constructors and one destructor as follows: The first constructor takes three parameters, int y, int m, int n; The second is the copy constructor that takes a DateV3 object as the parameter; The third is the default constructor that takes no parameter; The destructor takes no parameter. (3) Has overloaded operators: int operator-(DateV3 & oneDate); // return difference in days between the calling object and oneDate DateV3 operator+(int inc); // return a Date object that is inc days later than the calling object DateV3 operator-(int dec); // return a Date object that is dec days earlier than the calling object DateV3 operator++(); // overload the prefix ++ operator DateV3 operator++(int); // overload the postfix ++ operator friend ostream& operator<< (ostream& outputStream, DateV3& theDate); // overload the << operator Test class DateV3 in the main function as follows: (1) Declare and initialize an object to represent today, which should be the date that you work on this assignment. (2) Declare and initialize an object to represent your OWN birthday. (3) Express John’s birthday given John is 5 days older than yours. (4) Create Tom’s birthday by using the copy constructor, assuming Tom has the same birthday as you. (5) Display how many days have passed since your birth, John’s birth, and Tom’s birth, respectively. (6) Create an DateV3 object, someday, by cloning Tom’s birthday. Increment someday by the prefix operator ++ once, and by postfix operator ++ once. (7) Display someday, today, your birthday, John’s birthday, and Tom’s birthday. (8) Declare a DateV3 object to represent 28 February 2024, display it, apply the prefix ++ operator on it, display it again, and apply the postfix ++ operator on it and display it again. Hint: i) A good idea is to first design a function to compute the number of days that has passed since Year 1, Month 1, and Day 1, and then to use this function to compute the difference between two dates. ii) You can store the number of days for each of the 12 months in an integer array, which helps in counting the days and implementing the overloaded operators.

Write a C++ program that defines a class DateV2 that (1) Contains all the members in the class DateV1; Programming for Engineers C++ (2) Has two constructors as follows: One takes three parameters, int y, int m, int n; The other is the default constructor that takes no parameter (3) Has additional public member functions as follows: string getWeekDay(); // return the week day, for example, Sunday if day is 0, etc bool Leap(); // return if the year is leap int differFrom(DateV2& oneDate); // return the difference in days between the calling object // and the oneDate object void printDate(); // print the year, the month in English, the day, and the week day Test class DateV2 in the main function as follows: (1) Declare and set the objects today and tomorrow as in Problem 2. (2) Declare and initialize (by a constructor) an object to represent your OWN birthday. (3) Use the member function printDate to print today, tomorrow, and your birthday. (4) Output the weekday of today, tomorrow, and your own birthday. (5) Output how many days has passed since your birth (the difference between your birthday and today). Hint: i) We can use another string array to store the English name for week days (Sunday, Monday, through Saturday) ii) We know that it is Monday on Year 1, Month 1, and Day 1 iii) A good idea is to first design a function to compute the number of days that has passed since Year 1, Month 1, and Day 1, and then to use this function to compute the week day for a give date and to compute the difference between two dates. You can store the number of days for each of the 12 months in an integer array, which helps in counting the days.

C++ defines a class DateV3 with the following: private member variables: int year, month, day; Has three constructors and one destructor as follows: The first constructor takes three parameters, int y, int m, int n; The second is the copy constructor that takes a DateV3 object as the parameter; The third is the default constructor that takes no parameter; The destructor takes no parameter. (3) Has overloaded operators: int operator-(DateV3 & oneDate); // return difference in days between the calling object and oneDate DateV3 operator+(int inc); // return a Date object that is inc days later than the calling object DateV3 operator-(int dec); // return a Date object that is dec days earlier than the calling object DateV3 operator++(); // overload the prefix ++ operator DateV3 operator++(int); // overload the postfix ++ operator friend ostream& operator<< (ostream& outputStream, DateV3& theDate); // overload the << operator Test class DateV3 in the main function as follows: Declare and initialize an object to represent today, which should be the date that you work on this assignment.Declare and initialize an object to represent your OWN birthday.Express John’s birthday given John is 5 days older than yours. Create Tom’s birthday by using the copy constructor, assuming Tom has the same birthday as you. Display how many days have passed since your birth, John’s birth, and Tom’s birth, respectively. Create an DateV3 object, someday, by cloning Tom’s birthday. Increment someday by the prefix operator ++ once, and by postfix operator ++ once.Display someday, today, your birthday, John’s birthday, and Tom’s birthday. Declare a DateV3 object to represent 28 February 2024, display it, apply the prefix ++ operator on it, display it again, and apply the postfix ++ operator on it and display it again.

最新推荐

recommend-type

yolov5-face-landmarks-opencv

yolov5检测人脸和关键点,只依赖opencv库就可以运行,程序包含C++和Python两个版本的。 本套程序根据https://github.com/deepcam-cn/yolov5-face 里提供的训练模型.pt文件。转换成onnx文件, 然后使用opencv读取onnx文件做前向推理,onnx文件从百度云盘下载,下载 链接:https://pan.baidu.com/s/14qvEOB90CcVJwVC5jNcu3A 提取码:duwc 下载完成后,onnx文件存放目录里,C++版本的主程序是main_yolo.cpp,Python版本的主程序是main.py 。此外,还有一个main_export_onnx.py文件,它是读取pytorch训练模型.pt文件生成onnx文件的。 如果你想重新生成onnx文件,不能直接在该目录下运行的,你需要把文件拷贝到https://github.com/deepcam-cn/yolov5-face 的主目录里运行,就可以生成onnx文件。
recommend-type

setuptools-0.6c8-py2.5.egg

文件操作、数据分析和网络编程等。Python社区提供了大量的第三方库,如NumPy、Pandas和Requests,极大地丰富了Python的应用领域,从数据科学到Web开发。Python库的丰富性是Python成为最受欢迎的编程语言之一的关键原因之一。这些库不仅为初学者提供了快速入门的途径,而且为经验丰富的开发者提供了强大的工具,以高效率、高质量地完成复杂任务。例如,Matplotlib和Seaborn库在数据可视化领域内非常受欢迎,它们提供了广泛的工具和技术,可以创建高度定制化的图表和图形,帮助数据科学家和分析师在数据探索和结果展示中更有效地传达信息。
recommend-type

zigbee-cluster-library-specification

最新的zigbee-cluster-library-specification说明文档。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本:布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学,1996年。法语。NNT:电话:00345357HAL ID:电话:00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆,用于存放和传播科学研究论文,无论它们是否被公开。论文可以来自法国或国外的教学和研究机构,也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

实现实时数据湖架构:Kafka与Hive集成

![实现实时数据湖架构:Kafka与Hive集成](https://img-blog.csdnimg.cn/img_convert/10eb2e6972b3b6086286fc64c0b3ee41.jpeg) # 1. 实时数据湖架构概述** 实时数据湖是一种现代数据管理架构,它允许企业以低延迟的方式收集、存储和处理大量数据。与传统数据仓库不同,实时数据湖不依赖于预先定义的模式,而是采用灵活的架构,可以处理各种数据类型和格式。这种架构为企业提供了以下优势: - **实时洞察:**实时数据湖允许企业访问最新的数据,从而做出更明智的决策。 - **数据民主化:**实时数据湖使各种利益相关者都可
recommend-type

2. 通过python绘制y=e-xsin(2πx)图像

可以使用matplotlib库来绘制这个函数的图像。以下是一段示例代码: ```python import numpy as np import matplotlib.pyplot as plt def func(x): return np.exp(-x) * np.sin(2 * np.pi * x) x = np.linspace(0, 5, 500) y = func(x) plt.plot(x, y) plt.xlabel('x') plt.ylabel('y') plt.title('y = e^{-x} sin(2πx)') plt.show() ``` 运行这段
recommend-type

JSBSim Reference Manual

JSBSim参考手册,其中包含JSBSim简介,JSBSim配置文件xml的编写语法,编程手册以及一些应用实例等。其中有部分内容还没有写完,估计有生之年很难看到完整版了,但是内容还是很有参考价值的。
recommend-type

"互动学习:行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动,互动学习以行动为中心的强化学习学会互动,互动学习,以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授:智囊团论文联合主任菲利普·普雷教授,大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授,Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士,Deepmind对于那些及时看到自己错误的人...3谢谢你首先,我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔,"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲,你知道在这篇论文的(许多)错误中,你是我可以依
recommend-type

实现实时监控告警系统:Kafka与Grafana整合

![实现实时监控告警系统:Kafka与Grafana整合](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9tbWJpei5xcGljLmNuL21tYml6X2pwZy9BVldpY3ladXVDbEZpY1pLWmw2bUVaWXFUcEdLT1VDdkxRSmQxZXB5R1lxaWNlUjA2c0hFek5Qc3FyRktudFF1VDMxQVl3QTRXV2lhSWFRMEFRc0I1cW1ZOGcvNjQw?x-oss-process=image/format,png) # 1.1 Kafka集群架构 Kafka集群由多个称为代理的服务器组成,这
recommend-type

导入numpy库,创建两个包含9个随机数的3*3的矩阵,将两个矩阵分别打印出来,计算两个数组的点积并打印出来。(random.randn()、dot()函数)

可以的,以下是代码实现: ```python import numpy as np # 创建两个包含9个随机数的3*3的矩阵 matrix1 = np.random.randn(3, 3) matrix2 = np.random.randn(3, 3) # 打印两个矩阵 print("Matrix 1:\n", matrix1) print("Matrix 2:\n", matrix2) # 计算两个数组的点积并打印出来 dot_product = np.dot(matrix1, matrix2) print("Dot product:\n", dot_product) ``` 希望