Problem C: 从点到面

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 3317 Solved: 1854
[Submit][Status][Web Board]

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

append.cc,

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;
}

#include <iostream>
using namespace std;
class Point
{
public :
    double x_, y_;
    Point(double x=0):x_(x), y_(x)
    {
        cout<<"A point ("<<x_<<", "<<y_<<") is created!"<<endl;
    }
    Point(double x, double y):x_(x), y_(y)
    {
        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;
    }
    double getX(){return x_;}
    double getY(){return y_;}
    ~Point()
    {
        cout<<"A point ("<<x_<<", "<<y_<<") is erased!"<<endl;
    }
};
class Rectangle
{
public :
    Point leftTop, rightBottom;
    double x1, y1, x2, y2;
    Rectangle(double a, double b, double c, double d):leftTop(a,b), rightBottom(c,d)
    {
        x1=a; y1=b; x2=c; y2=d;
        cout<<"A rectangle ("<<a<<", "<<b<<") to ("<<c<<", "<<d<<") is created!"<<endl;
    }
    Point &getLeftTop()
    {
        return leftTop;
    }
    Point getRightBottome()//本题纯属胡闹，为了一个出现copy一个不出现copy一个加&一个不加&。
    {
        return rightBottom;
    }
    double getArea()
    {
        return(x1-x2)*(y2-y1);
    }
    ~Rectangle()
    {
         cout<<"A rectangle ("<<x1<<", "<<y1<<") to ("<<x2<<", "<<y2<<") is erased!"<<endl;
    }

};
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;
}