3-4 计算长方形的周长和面积

悠悠 2022-09-26 03:31 247阅读 0赞

### 3-4 计算长方形的周长和面积 ###

Time Limit: 1000MS Memory Limit: 65536KB

[Submit][] [Statistic][]

#### Problem Description ####

通过本题的练习可以掌握拷贝构造函数的定义和使用方法；

设计一个长方形类Rect，计算长方形的周长与面积。类中有私有数据成员Length(长)、Width(宽)，由具有缺省参数值的构造函数对其初始化，函数原型为：Rect(double Length=0, double Width=0); 再为其定义拷贝构造函数，形参为对象的常引用，函数原型为：Rect(const Rect &); 编写主函数，创建Rect对象r1初始化为长、宽数据，利用r1初始化另一个Rect对象r2，分别输出对象的长和宽、周长和面积。

要求： 创建对象 Rect r1(3.0,2.0),r2(r1);

#### Input ####

输入两个实数，中间用一个空格间隔；代表长方形的长和宽

#### Output ####

共有6行；  
分别输出r1的长和宽；r1的周长；r1的面积；r2的长和宽；r2的周长；r2的面积；注意单词与单词之间用一个空格间隔

#### Example Input ####

56 32

#### Example Output ####

the length and width of r1 is:56,32
    the perimeter of r1 is:176
    the area of r1 is:1792
    the length and width of r2 is:56,32
    the perimeter of r2 is:176
    the area of r2 is:1792

#### Hint ####

输入\-7.0 -8.0

输出

the length and width of r1 is:0,0

the perimeter of r1 is:0

the area of r1 is:0

the length and width of r2 is:0,0

the perimeter of r2 is:0

the area of r2 is:0

\#include <iostream>  
using namespace std;  
class Rect  
\{  
private:  
double length;  
double width;  
public:  
Rect(double x=0,double y =0)  
\{  
length = x;  
width = y;  
\}  
Rect(const Rect &b)  
\{  
length = b.length;  
width = b.width;  
\}  
const void display()  
\{  
cout << "the length and width of r1 is:" << length << ","<< width << endl;  
cout << "the perimeter of r1 is:" << (length + width) \* 2<<endl;  
cout << "the area of r1 is:" << length \* width << endl;  
\}  
void display1()  
\{  
cout << "the length and width of r2 is:" << length << ',' << width<< endl;  
cout << "the perimeter of r2 is:" << (length + width) \* 2 << endl;  
cout << "the area of r2 is:" << length \* width << endl;  
\}  
  
\};  
  
  
int main()  
\{  
double x,y;  
cin>>x>>y;  
if(x<0||y<0)  
\{  
x=0;  
y=0;  
\}  
Rect r1(x,y);  
Rect r2 = r1;  
r1.display();  
r2.display1();  
return 0;  
\}

[Submit]: http://www.sdutacm.org/onlinejudge2/index.php/Home/Contest/contestsubmit/cid/1818/pid/2673.html
[Statistic]: http://www.sdutacm.org/onlinejudge2/index.php/Home/Contest/conteststatus/cid/1818/pid/2673.html