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. Deep learning:四十六(DropConnect简单理解)

    和maxout(maxout简单理解)一样,DropConnect也是在ICML2013上发表的,同样也是为了提高Deep Network的泛化能力的,两者都号称是对Dropout(Dropout简单 ...

  2. Webydo:一款在线自由创建网站的 Web 应用

    Webydo 是一款专业的在线建站应用,使平面设计师可以创建和管理 HTML 网站,而无需编写代码.设计人员可以设计任何类型网站,只需要点击按钮,就能够发布先进的 HTML 网站. 你可以控制所有的设 ...

  3. BrowserSync - 浏览器同步测试工具

    背景: 之前在学gulp的时候,使用gulp-livereload来实时自动刷新页面省时开发,但一直比较难用,现在找到新的替代神器. 安装:   // 使用淘宝镜像会快些 npm install -g ...

  4. 白话Https

    本文试图以通俗易通的方式介绍Https的工作原理,不纠结具体的术语,不考证严格的流程.我相信弄懂了原理之后,到了具体操作和实现的时候,方向就不会错,然后条条大路通罗马.阅读文本需要提前大致了解对称加密 ...

  5. JavaScript之旅(三)

    JavaScript之旅(三) 三.函数 在JavaScript中,定义函数的方式如下: function abs(x) { ... return ...; } 如果没有return,返回结果为und ...

  6. visual studio 2015.3 downloads

    https://www.visualstudio.com/zh-hans/downloads/ visual studio 2015.3 downloads http://download.micro ...

  7. Error: pathspec '*' did not match any file(s) known to git.

    git切换分支报错 error: pathspec 'develop' did not match any file(s) known to git. 解决办法如下: plumm@MACY-PC MI ...

  8. css3很美的蟠桃动画

    查看效果:http://hovertree.com/texiao/css3/26/ 源码下载:http://hovertree.com/h/bjaf/ndhxgfkn.htm 效果图如下: 代码如下: ...

  9. 速战速决 (1) - PHP: 概述, 常量, 变量, 运算符, 表达式, 控制语句

    [源码下载] 速战速决 (1) - PHP: 概述, 常量, 变量, 运算符, 表达式, 控制语句 作者:webabcd 介绍速战速决 之 PHP 概述 常量 变量 运算符 表达式 控制语句 示例1. ...

  10. hibernate----1-N--jointable(人与地址)

    package com.ij34.dao; import java.util.HashMap; import java.util.HashSet; import java.util.Set; impo ...