实验9:Problem G: 克隆人来了!
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!
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
Zhang 20
Li 18
Zhao 99
Sample Output
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
-->
한국어<中文فارسیEnglishไทยAll Copyright Reserved 2010-2011SDUSTOJTEAMGPL2.02003-2011HUSTOJ ProjectTEAM
Anything about the Problems, Please Contact Admin:admin
#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
- 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 ...
- 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 ...
- 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 ...
- 【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 ...
- 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 ...
- Problem G: Keywords Search
Problem G: Keywords SearchTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 10 Solved: 6[Submit][Status] ...
- 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 ...
- 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 ...
- 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
- jQuery判断当前元素显示状态并控制元素的显示与隐藏
1.jQuery判断一个元素当前状态是显示还是隐藏 $("#id").is(':visible'); //true为显示,false为隐藏 $("#id") ...
- vs切换当前编辑文件时自动定位目录树
在编辑区,切换当前编辑文件时(单击.cpp或.h文件选项卡),"解决方案资源管理器"目录树会自动定位当前编辑的文件,并以灰色标识,当一个解决方案中的工程数目数目很多,每个工程下面又 ...
- Testing - 测试基础 - 探索
定义 探索性测试(Exploratory Testing)是一种自由的软件测试风格,强调测试人员同时展开测试学习,测试设计,测试执行和测试结果评估等活动,以持续优化测试工作. 其特征有:即兴发挥,快速 ...
- [git]merge和rebase的区别
前言 我从用git就一直用rebase,但是新的公司需要用merge命令,我不是很明白,所以查了一些资料,总结了下面的内容,如果有什么不妥的地方,还望指正,我一定虚心学习. merge和rebase ...
- Android SlidingMenu 仿网易新闻客户端布局
前面两篇文章中的SlidingMenu都出现在左侧,今天来模仿一下网易新闻客户端左右两边都有SlidingMenu的效果,以下是网易新闻客户端效果: 不扯闲话了,直接进入正题吧 frame_conte ...
- Android性能测试工具(一)之Emmagee
Android性能测试工具(一) 之Emmagee Emmagee是监控指定被测应用在使用过程中占用机器的CPU.内存.流量资源的性能测试小工具. 支持SDK:Android2.2以及以上版本 Emm ...
- Windows Azure Traffic Manager (5) Traffic Manager Overview
<Windows Azure Platform 系列文章目录> 笔者默默地看了一下之前写的Traffic Manager内容,已经差不多是3年前的文章了.现在Azure Traffic M ...
- The SQL Server Service Broker for the current database is not enabled, and as a result query notifications are not supported.
当Insus.NET尝试解决此问题<When using SqlDependency without providing an options value, SqlDependency.Star ...
- .NET Core配置文件加载与DI注入配置数据
.NET Core配置文件 在以前.NET中配置文件都是以App.config / Web.config等XML格式的配置文件,而.NET Core中建议使用以JSON为格式的配置文件,因为使用起来更 ...
- webstorage[html5的本地数据处理]
1.webStorage是什么? webStorage是html5中用于本地化存储的一种方式,而在之前呢我们是用cookie的存储方式处理; 2.那它们之间的区别是什么? Ⅰ.cookie存在的问题: ...