Problem D: 从点到面

Description

一个矩形可以由左上角和右下角的顶点而唯一确定。现在请定义两个类：Point和Rectangle。

其中Point类有x和y两个属性（均为int类型），表示二维空间内一个点的横纵坐标，并具有相应的构造函数、析构函数和拷贝构造函数。此外，还有getX()和getY()方法用以得到一个点的坐标值。

Rectangle类有leftTop和rightBottom两个属性（均为Point类的对象），表示一个矩形的左上角和右下角的两个点，并具有相应的构造函数、析构函数。此外，还有getLeftTop()、getRightBottom()方法用于获取相应的左上角点、右下角点，getArea()方法用以获取面积。

Input

输入有多行。

第一行是一个正整数M，表示后面有M个测试用例。

每个测试用例占一行，包括4个正整数，分别为左上角的横坐标、纵坐标，右下角的横坐标、纵坐标。

注意：

1.请根据输出样例判断两个类中相应方法的书写方法。

2. 假定屏幕的左下角为坐标原点。

Output

输出见样例。

Sample Input

1

10 10 20 0

Sample Output

A point (10, 10) is created!

A point (20, 0) is created!

A rectangle (10, 10) to (20, 0) is created!

Area: 100

Left top is (10, 10)

A point (20, 0) is copied!

A point (20, 0) is copied!

Right bottom is (20, 0)

A point (20, 0) is erased!

A point (20, 0) is erased!

A rectangle (10, 10) to (20, 0) is erased!

A point (20, 0) is erased!

A point (10, 10) is erased!

HINT

Append Code

#include<iostream>

using namespace std;

class Point

{

private :

     int x,y;

public :

     Point( int a, int b){x=a,y=b;cout<< "A point (" <<x<< ", " <<y<< ") is created!\n" ;}

     ~Point(){cout<< "A point (" <<x<< ", " <<y<< ") is erased!\n" ;}

     Point( const Point &p){x=p.x,y=p.y;cout<< "A point (" <<x<< ", " <<y<< ") is copied!\n" ;}

     int getX(){ return x;}

     int getY(){ return y;}

};

class Rectangle

{

private :

     Point leftTop,rightBottom; //leftTop.x(a),leftTop.y(b),rightBottom.x(c),rightBottom.y(d)

public :

     Rectangle( int a, int b, int c, int d):leftTop(a,b),rightBottom(c,d){cout<< "A rectangle (" <<leftTop.getX()<< ", " <<leftTop.getY()<< ") to (" <<rightBottom.getX()<< ", " <<rightBottom.getY()<< ") is created!\n" ;} //

     //Rectangle(Point a,Point b):leftTop(a),rightBottom(b){cout<<"A rectangle ("<<leftTop.getX()<<", "<<leftTop.getY()<<") to ("<<rightBottom.getX()<<", "<<rightBottom.getY()<<") is created!\n";}

     ~Rectangle(){cout<< "A rectangle (" <<leftTop.getX()<< ", " <<leftTop.getY()<< ") to (" <<rightBottom.getX()<< ", " <<rightBottom.getY()<< ") is erased!\n" ;}

     Point &getLeftTop(){ return leftTop;}

     Point getRightBottome(){ return rightBottom;}

     int getArea(){ return (rightBottom.getX()-leftTop.getX())*(leftTop.getY()-rightBottom.getY());}

};

int main()

{

     int cases;

     int x1, y1, x2, y2;

 

     cin>>cases;

     for ( int i = 0; i < cases; i++)

     {

         cin>>x1>>y1>>x2>>y2;

         Rectangle rect(x1,y1,x2,y2);

         cout<< "Area: " <<rect.getArea()<<endl;

         cout<< "Left top is (" <<rect.getLeftTop().getX()<< ", " <<rect.getLeftTop().getY()<< ")" <<endl;

         cout<< "Right bottom is (" <<rect.getRightBottome().getX()<< ", " <<rect.getRightBottome().getY()<< ")" <<endl;

     }

     return 0;

}

转载于:https://www.cnblogs.com/TogetherLaugh/p/6544703.html