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. $.when(deferreds)

    作者:禅楼望月(http://www.cnblogs.com/yaoyinglong ) 1 引子 上一篇博文中介绍的Deferred,它表示一个延迟对象.但是很多时候,我们需要在多个延迟对象(异步代 ...

  2. Android中自定义属性(attrs.xml,TypedArray的使用)

    做Android布局是件很享受的事,这得益于他良好的xml方式.使用xml可以快速有效的为软件定义界面.可是有时候我们总感觉官方定义的一些基本组件不够用,自定义组件就不可避免了.那么如何才能做到像官方 ...

  3. URL格式

    URL由三部分组成:资源类型.存放资源的主机域名.资源文件名. URL的一般语法格式为: (带方括号[]的为可选项): protocol :// hostname[:port] / path / [; ...

  4. 我如何调优SQL Server查询

    我是个懒人,我只想干尽可能少的活.当我干活的时候我不想太多.是,你没看错,这看起来很糟糕,作为一个DBA这很不合格.但在今天的文章里,我想给你展示下,当你想对特定查询创建索引设计时,你如何把你的工作和 ...

  5. php多进程处理

    php多进程处理 往往我们会碰到一个情况,需要写一个脚本,这个脚本要处理的数据量极大,单进程处理脚本非常慢,那么这个时候就会想到使用多进程或者多线程的方式了. 我习惯使用多进程的方式,php中使用多进 ...

  6. WebGL实现HTML5的3D贪吃蛇游戏

    js1k.com收集了小于1k的javascript小例子,里面有很多很炫很酷的游戏和特效,今年规则又增加了新花样,传统的classic类型基础上又增加了WebGL类型,以及允许增加到2K的++类型, ...

  7. HTML5的五种客户端离线存储方案

    最近折腾HTML5游戏需要离线存储功能,便把目前可用的几种HTML5存储方式研究了下,基于HT for Web写了个综合的实例,分别利用了Cookie.WebStorage.IndexedDB以及Fi ...

  8. 基于HT for Web 3D技术快速搭建设备面板

    以真实设备为模型,搭建出设备面板,并实时获取设备运行参数,显示在设备面板上,这相比于纯数值的设备监控系统显得更加生动直观.今天我们就在HT for Web的3D技术上完成设备面板的搭建. 我们今天模拟 ...

  9. Java魔法堂:枚举类型详解

    一.前言 Java的枚举类型相对C#来说具有更灵活可配置性,Java的枚举类型可以携带更多的信息. // C# enum MyColor{ RED = , BLUE = } Console.Write ...

  10. 使用Uploadify(UploadiFive)多文件上传控件遇到的坑

    最近项目中需要实现多文件上传功能,于是结合需求最终选择了Uploadify这一款控件来实现.相比其他控件,Uploadify具有简洁的界面,功能API基本可以解决大多数需求,又是基于jquery的,配 ...