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SDUST 点在圆内吗?

作者:互联网

Description

定义一个Point类和Circle类,用于判断给定的一系列的点是否在给定的圆内。

其中,Point类:

1.有2个成员x和y,分别为其横坐标和纵坐标;1个静态成员numOfPoints,用于计算生成的点的个数。

2.具有构造函数、析构函数和拷贝构造函数,具体格式输出根据样例自行判断。

3. 具有静态方法int getNumOfPoints(),用于返回numOfPoints的值。

4. 具有int getX()和int getY()方法,用于获取横坐标和纵坐标。

Circle类:

1. 拥有Point类的对象center,表示圆心坐标。拥有radius对象,表示圆的半径;1个静态成员numOfCircles,用于指示生成了多少个圆对象。

2. 具有构造函数、析构函数和拷贝构造函数,具体格式根据样例自行判断。

3.具有静态方法int getNumOfCircles(),返回numOfCircles的值。

4. 具有getCenter()方法,返回圆心坐标。注意:根据输出结果判断返回值类型。

5. 具有bool pointInCircle(Point &)方法,用于判断给定的点是否在当前圆内。是则返回true,否则返回false。

Input

输入分多行。

第一行M>0,表示有M个测试用例。

每个测试用例又包括多行。第1行包含3个整数,分别为一个圆的横坐标、纵坐标和半径。第2行N>0,表示之后又N个点,每个点占一行,分别为其横坐标和纵坐标。

所有输入均为整数,且在int类型范围内。

Output

输出见样例。注意:在圆的边上的点,不在圆内。

Sample Input

2

0 0 10

3

2 2

11 11

10 0

1 1 20

3

2 2

1 1

100 100

Sample Output

The Point (0, 0) is created!Now, we have 1 points.

The Point (1, 1) is created! Now, we have 2 points.

A circle at (1, 1) and radius 1 is created! Now, we have 1 circles.

We have 2 points and 1 circles now.

The Point (0, 0) is created! Now, we have 3 points.

A Point (0, 0) is copied! Now, we have 4 points.

A Point (0, 0) is copied! Now, we have 5 points.

A circle at (0, 0) and radius 10 is created! Now, we have 2 circles.

A Point (0, 0) is erased! Now, we have 4 points.

The Point (2, 2) is created! Now, we have 5 points.

(2, 2) is in the circle at (0, 0).

The Point (11, 11) is created! Now, we have 6 points.

(11, 11) is not in the circle at (0, 0).

The Point (10, 0) is created! Now, we have 7 points.

(10, 0) is not in the circle at (0, 0).

A Point (0, 0) is erased! Now, we have 6 points.

A circle at (0, 0) and radius 10 is erased! Now, we have 1 circles.

A Point (0, 0) is erased! Now, we have 5 points.

The Point (1, 1) is created! Now, we have 6 points.

A Point (1, 1) is copied! Now, we have 7 points.

A Point (1, 1) is copied! Now, we have 8 points.

A circle at (1, 1) and radius 20 is created! Now, we have 2 circles.

A Point (1, 1) is erased! Now, we have 7 points.

The Point (2, 2) is created! Now, we have 8 points.

(2, 2) is in the circle at (1, 1).

The Point (1, 1) is created! Now, we have 9 points.

(1, 1) is in the circle at (1, 1).

The Point (100, 100) is created! Now, we have 10 points.

(100, 100) is not in the circle at (1, 1).

A Point (1, 1) is erased! Now, we have 9 points.

A circle at (1, 1) and radius 20 is erased! Now, we have 1 circles.

A Point (1, 1) is erased! Now, we have 8 points.

We have 8 points, and 1 circles.

A circle at (1, 1) and radius 1 is erased!Now, we have 0 circles.

A Point (1, 1) is erased! Now, we have 7 points.

A Point (0, 0) is erased! Now, we have 6 points.

HINT

 

#include<iostream>
using namespace std;
class Point
{
    friend class Circle;
private:
    int x,y;
    static int numOfPoints;
public:
    Point(int a=0,int b=0):x(a),y(b)
    {
        numOfPoints++;
        cout<<"The Point ("<<x<<", "<<y<<") is created! Now, we have "<<numOfPoints<<" points."<<endl;
    }
    Point(const Point&p)
    {
        numOfPoints++;
        x=p.x;
        y=p.y;
        cout<<"A Point ("<<x<<", "<<y<<") is copied! Now, we have "<<numOfPoints<<" points."<<endl;
    }
    static int getNumOfPoints()
    {
        return numOfPoints;
    }
    int getX()
    {
        return x;
    }
    int getY()
    {
        return y;
    }
    ~Point()
    {
        numOfPoints--;
        cout<<"A Point ("<<x<<", "<<y<<") is erased! Now, we have "<<numOfPoints<<" points."<<endl;
    }
};
class Circle
{
private:
    Point center;
    int radius;
    static int numOfCircles;
public:
    Circle(int a,int b,int c):center(a,b),radius(c)
    {
        numOfCircles++;
        cout<<"A circle at ("<<center.x<<", "<<center.y<<") and radius "<<radius<<" is created! Now, we have "<<numOfCircles<<" circles."<<endl;
    }
    Circle(Point p,int r):center(p),radius(r)
    {
        numOfCircles++;
        cout<<"A circle at ("<<center.x<<", "<<center.y<<") and radius "<<radius<<" is created! Now, we have "<<numOfCircles<<" circles."<<endl;
    }
    Circle(const Circle &c)
    {
        numOfCircles++;
        center.x=c.center.x;
        center.y=c.center.y;
        radius=c.radius;
        cout<<"A circle at ("<<center.x<<", "<<center.y<<") and radius "<<radius<<" is created! Now, we have "<<numOfCircles<<" circles."<<endl;
    }
    static int getNumOfCircles()
    {
        return numOfCircles;
    }
    Point &getCenter()
    {
        return center;
    }
    ~Circle()
    {
        numOfCircles--;
        cout<<"A circle at ("<<center.x<<", "<<center.y<<") and radius "<<radius<<" is erased! Now, we have "<<numOfCircles<<" circles."<<endl;
    }
    bool pointInCircle(Point &p)
    {
        if((p.x-center.x)*(p.x-center.x)+(p.y-center.y)*(p.y-center.y)<radius*radius) return true;
        else return false;
    }

};
int Point::numOfPoints=0;
int Circle::numOfCircles=0;
 
int main()
{
    int cases,num;
    int x, y, r, px, py;
    Point aPoint(0,0), *bPoint;
    Circle aCircle(1,1,1);
    cin>>cases;
    cout<<"We have "<<Point::getNumOfPoints()<<" points and "<<Circle::getNumOfCircles()<<" circles now."<<endl;
    for (int i = 0; i < cases; i++)
    {
        cin>>x>>y>>r;
        bPoint = new Point(x,y);
        Circle circle(*bPoint, r);
        cin>>num;
        for (int j = 0; j < num; j++)
        {
            cin>>px>>py;
            if (circle.pointInCircle(*(new Point(px, py))))
            {
                cout<<"("<<px<<", "<<py<<") is in the circle at (";
                cout<<circle.getCenter().getX()<<", "<<circle.getCenter().getY()<<")."<<endl;
            }
            else
            {
                cout<<"("<<px<<", "<<py<<") is not in the circle at (";
                cout<<circle.getCenter().getX()<<", "<<circle.getCenter().getY()<<")."<<endl;
            }
        }
        delete bPoint;
    }
    cout<<"We have "<<Point::getNumOfPoints()<<" points, and "<<Circle::getNumOfCircles()<<" circles."<<endl;
    return 0;
}

 

标签:erased,Point,created,SDUST,points,圆内,circle,Now
来源: https://blog.csdn.net/lx18303916/article/details/88353531