写出enter your name. alice hello alice, do you want to repeat this program? yes enter your name. bob hello bob, do you want to repeat this program? yes enter your name. charlie hello charlie, do you want to repeat this program? no goodbye, charlie! 的c++代码


以下是一个简单的 C++ 代码示例,实现了你描述的功能: ```cpp #include <iostream> #include <string> int main() { std::string name; std::cout << "enter your name." << std::endl; std::cin >> name; std::cout << "hello " << name << ", do you want to repeat this program?" << std::endl; std::string answer; std::cin >> answer; while (answer == "yes") { std::cout << "enter your name." << std::endl; std::cin >> name; std::cout << "hello " << name << ", do you want to repeat this program?" << std::endl; std::cin >> answer; } if (answer == "no") { std::cout << "goodbye, " << name << "!" << std::endl; } return 0; } ``` 这段代码首先提示用户输入名字,然后询问是否重复程序。如果回答是"yes",则再次执行输入名字和询问的过程。如果回答是"no",则输出再见的消息。 [2024-10-22 08:36:28 | AI写代码神器 | 267点数解答]
- 写出enter your name. alice hello alice, do you want to repeat this program? yes enter your name. bob hello bob, do you want to repeat this program? yes enter your name. charlie hello charlie, do you want to repeat this program? no goodbye, charlie! 的c++代码(267点数解答 | 2024-10-22 08:36:28)146
- 使用下列代码创建数据框,完成以下的填空。 import numpy as np import pandas as pd data = {'animal': ['cat', 'cat', 'snake', 'dog', 'dog', 'cat', 'snake', 'cat', 'dog', 'dog'], 'age': [2.5, 3, 0.5, np.nan, 5, 2, 4.5, np.nan, 7, 3], 'visits': [1, 3, 2, 3, 2, 3, 1, 1, 2, 1], 'priority': ['yes', 'yes', 'no', 'yes', 'no', 'no', 'no', 'yes', 'no', 'no']} labels = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'] 1、使用data为数据,labels为索引,创建数据框df 2、取出 df 的 animal 和 age 列 填空1:df[________](317点数解答 | 2025-04-14 18:32:16)111
- 要求:创建一个对象 const person = {name: 'alice', age: 25, city: 'new york'},使用解构赋值从对象中提取 name 和 age,然后使用模板字符串创建一句话,例如“姓名是 alice,年龄是 25。” const person = {name: 'alice', age: 25, city: 'new york'}; // 使用解构赋值和模板字符串(138点数解答 | 2024-10-29 09:40:13)125
- 解构赋值和模板字符串 要求:创建一个对象 const person = {name: 'alice', age: 25, city: 'new york'},使用解构赋值从对象中提取 name 和 age,然后使用模板字符串创建一句话,例如“姓名是 alice,年龄是 25。” const person = {name: 'alice', age: 25, city: 'new york'}; // 使用解构赋值和模板字符串(64点数解答 | 2024-10-29 21:04:59)166
- 2.1 description john is learning a new programming language called a++. having just mastered loops, he is excitedly writing many programs that contain only loop structures. however, he does not know how to compute the time complexity of his programs. so he turns to you for help. what you need to do is to write a program to calculate the time complexity for each program that john writes. the loop structure in a++ is as follows: f i x y ...//code block to be executed e here ”f i x y” indicates cre(493点数解答 | 2024-10-24 03:51:13)164
- 2.1 description john is learning a new programming language called a++. having just mastered loops, he is excitedly writing many programs that contain only loop structures. however, he does not know how to compute the time complexity of his programs. so he turns to you for help. what you need to do is to write a program to calculate the time complexity for each program that john writes. the loop structure in a++ is as follows: f i x y ...//code block to be executed e here ”f i x y” indicates cre(720点数解答 | 2024-10-24 03:51:50)135
- 题目描述 输入两个整数 a , b a,b 1. 1. 对 a , b a,b之间个位数是 3 3的数字求和并输出。 2. 2. 判断这个和是不是3的倍数,如果这个数字是3的倍数,请输出: Y E S YES,否则的话输出: N O NO。 输入格式 输入包括一行,包含两个整数 a , b a,b,数字之间用空格隔开。 输出格式 输出包括两行 第一行为 a ∼ b a∼b 之间 个位数是 3 3 的数字和。 第二行 如果这个数字是3的倍数,请输出: Y E S YES,否则的话输出: N O NO。 input1 复制 1 10 output1 复制 3 YES input2 复制 4 21 output2 复制 13 NO 样例解释 对于样例 1 1: 1 ∼ 10 1∼10 之间个位数是 3 3数字有 1 1 个,是 3 3,所以第一行输出 3 3,它是 3 3的倍数,所以第二行输出: Y E S YES。 对于样例 2 2: 4 ∼ 21 4∼21 之间个位数是 3 3数字有 1 1 个,是 13 13,所以第一行输出 13 13,它不是 3 3(358点数解答 | 2025-07-12 21:48:22)72
- 使用下列代码创建数据框,完成以下的填空题。 import numpy as np import pandas as pd index = pd.Index(data=["Tom", "Bob", "Mary", "James", "Andy", "Alice"], name="name") data = { "age": [18, 30, np.nan, 40, np.nan, 30], "city": ["Bei Jing ", "Shang Hai ", "Guang Zhou", "Shen Zhen", np.nan, " "], "income": [None, 11, 22,33, np.nan, 55], "sex": [None, "male", "female", "male", np.nan, "male"], "birth": ["2000-02-10", "1988-10-17", None, "1978-08-08", np.nan, "1988-10-17"] } (1)使用data为数据,ind(563点数解答 | 2025-04-14 18:30:52)80
- is it a cat? time limit per test 2 seconds memory limit per test 256 megabytes you were walking down the street and heard a sound. the sound was described by the string s consisting of lowercase and uppercase latin characters. now you want to find out if the sound was a cat meowing. for the sound to be a meowing, the string can only contain the letters 'm', 'e', 'o' and 'w', in either uppercase or lowercase. also: string must start with non-empty sequence consisting only of characters 'm' o(180点数解答 | 2024-12-17 01:45:05)191
- 用html创建一个对象 const person = {name: 'alice', age: 25, city: 'new york'} ,使用解构赋值从对象中提取 name 和 age ,然后使用模板字符串创建一句话,例如“姓名是 alice,年龄是 25。”(159点数解答 | 2024-10-26 23:27:20)113
- 给以下代码进行注释: for i=0 to selection.count do ( if i!=0 then ( if superclassof selection[i]==geometryclass do ( tomesh = snapshotasmesh selection[i]; if tomesh.numverts!=0 then ( sz=calculatevolumeandcenterofmass selection[i]; for j=1 to cdsz.count do ( xlc = xlx.application.cells (i+1) j; cdsy=finditem quansz cdsz[j];--菜单索引 xlc.value=sz[cdsy]; ) ) ) ) else( for j=1 to cdsz.count do ( xlc = xlx.application(386点数解答 | 2024-10-16 19:29:32)166
- 给以下代码进行注释: for i=0 to selection.count do ( if i!=0 then ( if superclassof selection[i]==geometryclass do ( tomesh = snapshotasmesh selection[i]; if tomesh.numverts!=0 then ( sz=calculatevolumeandcenterofmass selection[i]; for j=1 to cdsz.count do ( xlc = xlx.application.cells (i+1) j; cdsy=finditem quansz cdsz[j];--菜单索引 xlc.value=sz[cdsy]; ) ) ) ) else( for j=1 to cdsz.count do ( xlc = xlx.application(698点数解答 | 2024-10-16 19:33:34)198