想要输出""的话:

cout<<"A person whose name is \""<<name<<"\" and age is "<<age<<" is created!"<<endl;
Home Web Board ProblemSet Standing Status Statistics
 

Problem G: 克隆人来了!

Problem G: 克隆人来了!

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 448  Solved: 263
[Submit][Status][Web Board]

Description

克隆技术飞速发展,克隆人已经成为现实了!!所以,现在由你来编写一个Person类,来模拟其中的克隆过程。这个类具有2个属性:name——姓名(char*类型),和age——年龄(int类型)。

该类具有无参构造函数(人名为“no name”,年龄是0)、带参数构造函数、拷贝构造函数以及析构函数外,还有以下3个成员函数:

1. void Person::showPerson():按照指定格式显示人的信息。

2. Person& Person::setName(char *):设定人的姓名。

3. Person& Person::setAge(int):设定人的年龄。

Input

输入分多行,第一行是一个正整数N,表示其后有N行输入。每行分两部分:第一部分是一个没有空白符的字符串,表示一个人的姓名;第二部分是一个正整数,表示人的年龄。

Output

呃~比较复杂,见样例吧!注意:要根据样例编写相应函数中的输出语句,注意格式哦!

Sample Input

3
Zhang 20
Li 18
Zhao 99

Sample Output

A person whose name is "no name" and age is 0 is created!
A person whose name is "Tom" and age is 16 is created!
A person whose name is "Tom" and age is 16 is cloned!
A person whose name is "Zhang" and age is 20 is created!
This person is "Zhang" whose age is 20.
A person whose name is "Zhang" and age is 20 is erased!
A person whose name is "Li" and age is 18 is created!
This person is "Li" whose age is 18.
A person whose name is "Li" and age is 18 is erased!
A person whose name is "Zhao" and age is 99 is created!
This person is "Zhao" whose age is 99.
A person whose name is "Zhao" and age is 99 is erased!
This person is "Zhao" whose age is 18.
This person is "no name" whose age is 0.
A person whose name is "Zhao" and age is 18 is erased!
A person whose name is "Tom" and age is 16 is erased!
A person whose name is "no name" and age is 0 is erased!

HINT

注意:输出中有“”!

Append Code

[Submit][Status][Web Board]

#include<iostream>
using namespace std;
class Person{
public:
char* name;
int age;
Person():name("no name"),age(){cout<<"A person whose name is \""<<name<<"\" and age is "<<age<<" is created!"<<endl;}
Person(char* n,int a):name(n),age(a){cout<<"A person whose name is \""<<name<<"\" and age is "<<age<<" is created!"<<endl;}
Person(const Person& p){name=p.name;age=p.age;cout<<"A person whose name is \""<<name<<"\" and age is "<<age<<" is cloned!"<<endl;}
~Person(){cout<<"A person whose name is \""<<name<<"\" and age is "<<age<<" is erased!"<<endl;} void showPerson(){cout<<"This person is \""<<name<<"\" whose age is "<<age<<"."<<endl;}
Person& setName(char* n){name=n;return *this;}
Person& setAge(int a){age=a;return *this;}
}; int main()
{
int cases;
char str[];
int age; Person noname, Tom("Tom", ), anotherTom(Tom);
cin>>cases;
for (int ca = ; ca < cases; ca++)
{
cin>>str>>age;
Person newPerson(str, age);
newPerson.showPerson();
}
anotherTom.setName(str).setAge();
anotherTom.showPerson();
noname.showPerson();
return ;
}

实验9:Problem G: 克隆人来了!的更多相关文章

  1. 实验12:Problem G: 强悍的矩阵运算来了

    这个题目主要是乘法运算符的重载,卡了我好久,矩阵的乘法用3个嵌套的for循环进行,要分清楚矩阵的乘法结果是第一个矩阵的行,第二个矩阵的列所组成的矩阵. 重载+,*运算符时,可以在参数列表中传两个矩阵引 ...

  2. 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem G: Check The Check(模拟国际象棋)

    Problem G: Check The Check Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 10  Solved: 3[Submit][Statu ...

  3. The Ninth Hunan Collegiate Programming Contest (2013) Problem G

    Problem G Good Teacher I want to be a good teacher, so at least I need to remember all the student n ...

  4. 【贪心+中位数】【新生赛3 1007题】 Problem G (K)

    Problem G Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Sub ...

  5. Problem G: If We Were a Child Again

    Problem G: If We Were a Child AgainTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 18 Solved: 14[Submi ...

  6. Problem G: Keywords Search

    Problem G: Keywords SearchTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 10 Solved: 6[Submit][Status] ...

  7. BZOJ4977 八月月赛 Problem G 跳伞求生 set 贪心

    欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - BZOJ4977 - 八月月赛 Problem G 题意 小明组建了一支由n名玩家组成的战队,编号依次为1到n ...

  8. Western Subregional of NEERC, Minsk, Wednesday, November 4, 2015 Problem G. k-palindrome dp

    Problem G. k-palindrome 题目连接: http://opentrains.snarknews.info/~ejudge/team.cgi?SID=c75360ed7f2c7022 ...

  9. ZOJ 4010 Neighboring Characters(ZOJ Monthly, March 2018 Problem G,字符串匹配)

    题目链接  ZOJ Monthly, March 2018 Problem G 题意  给定一个字符串.现在求一个下标范围$[0, n - 1]$的$01$序列$f$.$f[x] = 1$表示存在一种 ...

随机推荐

  1. 【环境配置】Linux环境下下载、配置java环境、安装eclipse、建立eclipse快捷方式详解

    一.首先是下载Java JDK 到目前为止的最新版本为(jdk1.8.0_60),有两种方式进行下载: 1.使用shell来进行下载,可使用如下命令直接进行下载: wget --no-check-ce ...

  2. [New Portal]Windows Azure Virtual Machine (11) 在本地使用Hyper-V制作虚拟机模板,并上传至Azure (1)

    <Windows Azure Platform 系列文章目录> 本章介绍的内容是将本地Hyper-V的VHD,上传到Azure数据中心,作为自定义的虚拟机模板. 注意:因为在制作VHD的最 ...

  3. 利用开源软件strongSwan实现支持IKEv2的企业级IPsec VPN,并结合FreeRadius实现AAA协议(下篇)

    续篇—— 利用开源软件strongSwan实现支持IKEv2的企业级IPsec VPN,并结合FreeRadius实现AAA协议(上篇) 上篇文章写了如何构建一个支持IKEv2的VPN,本篇记录的是如 ...

  4. Math.ceil(a/b)结果出错--原因是a和b不是double

    脑袋短路.连续测试几次发现Math.ceil(188/20)==9; 忍无可忍,突然发现是int问题,顺着表达式走一遍,188/20==9,然后再向上取整.脑袋僵化了.看来一直做简单的不动脑筋的工作, ...

  5. 最简单的pagging插件

    <html> <head> <title>jQuery Easy-Paging Test</title> </head> <body& ...

  6. 轻松实现localStorage本地存储

    相信大家都知道HTML5提供了localStorage和sessionStorage两个新功能,基于这两个功能我们可以实现web资源的离线和会话存储,如果你现在还在用Cookie来临时存储网络资源的话 ...

  7. 30天C#基础巩固------了解委托,string练习

    ---->了解委托.     生活中的例子:我要打官司,我需要找一个律师,法庭上面律师为当事人辩护,它真正执行的是当事人的陈词,这时律师 就相当于一个委托对象.当事人则委托律师为自己辩解.    ...

  8. EF更新,数据库值变化,前台页面并不变化,刷新也不变化,重新运行程序则变化----开发中遇到的问题(已解决)

    首先说一下我遇到这个情况的代码情景,首先上错误代码 UserInfo userInfo = Session["UserInfo"] as UserInfo; ); 这段代码所呈现的 ...

  9. PHP多种形式发送邮件

    1. 使用 mail() 函数 没什么好讲的,就是使用系统自带的smtp系统来发送,一般是使用sendmail来发.这个按照各个系统不同而定.使用参考手册. 2. 使用管道的形式 昨天刚测试成功,使用 ...

  10. ASP.NET MVC中使用Dropzone.js实现图片的批量拖拽上传

    说在前面 最近在做一个MVC相册的网站(这里),需要批量上传照片功能,所以就在网上搜相关的插件,偶然机会发现Dropzone.js,试用了一下完全符合我的要求,而且样式挺满意的,于是就在我的项目中使用 ...