1777 Problem 从点到面

题目描述

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。

输入

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。

输出

输出见样例。

样例输入

1 10 10 20 0

样例输出

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!

提示

Append Code

append.cc,

#include<iostream>
#include<cstring>
#include<cmath>
#include<iomanip>
using namespace std;
class Point
{
private:
    int x;
    int y;
public:
    Point(int x1,int y1):x(x1),y(y1){cout<<"A point ("<<x<<", "<<y<<") is created!"<<endl;}
    Point(const Point &p)
    {
        x=p.x;
        y=p.y;
        cout<<"A point ("<<x<<", "<<y<<") is copied!"<<endl;
    }
    ~Point(){cout<<"A point ("<<x<<", "<<y<<") is erased!"<<endl;}
    int getX(){return x;}
    int getY(){return y;}
};
class Rectangle
{
private:
    Point leftTop;
    Point rightBottom;
public:
    Rectangle(int x1,int y1,int x2,int y2):leftTop(x1,y1),rightBottom(x2,y2){
    cout<<"A rectangle ("<<x1<<", "<<y1<<") to ("<<x2<<", "<<y2<<") is created!"<<endl;
    }
    ~Rectangle(){cout<<"A rectangle ("<<leftTop.getX()<<", "<<leftTop.getY()<<") to ("<<rightBottom.getX()<<", "<<rightBottom.getY()<<") is erased!"<<endl;}

    Point &getLeftTop(){return leftTop;}
    Point getRightBottome(){return rightBottom;}
    int getArea(){int S=(rightBottom.getX()-leftTop.getX())*(leftTop.getY()-rightBottom.getY());return S;}

};
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;
}