Problem E: 点在圆内吗？

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

Append Code

#include<iostream>

using namespace std;

class Point

{

private :

     int x,y;

     static int numOfPoints;

public :

     Point( int a, int b):x(a),y(b){numOfPoints++;cout<< "The Point (" <<x<< ", " <<y<< ") is created! Now, we have " <<numOfPoints<< " points.\n" ;}

     ~Point(){numOfPoints--;cout<< "A Point (" <<x<< ", " <<y<< ") is erased! Now, we have " <<numOfPoints<< " points.\n" ;}

     Point( const Point &q):x(q.x),y(q.y){numOfPoints++;cout<< "A Point (" <<x<< ", " <<y<< ") is copied! Now, we have " <<numOfPoints<< " points.\n" ;}

     static int getNumOfPoints(){ return numOfPoints;}

     int getX(){ return x;}

     int getY(){ return y;}

};

int Point::numOfPoints=0;

class Circle

{

private :

     Point center;

     int radius;

     static int numOfCircles;

public :

     Circle(Point b, int a):center(b),radius(a){numOfCircles++;cout<< "A circle at (" <<center.getX()<< ", " <<center.getY()<< ") and radius " <<radius<< " is created! Now, we have " <<numOfCircles<< " circles.\n" ;}

     Circle( int a, int b, int c):center(a,b),radius(c){numOfCircles++;cout<< "A circle at (" <<center.getX()<< ", " <<center.getY()<< ") and radius " <<radius<< " is created! Now, we have " <<numOfCircles<< " circles.\n" ;}

     ~Circle(){numOfCircles--;cout<< "A circle at (" <<center.getX()<< ", " <<center.getY()<< ") and radius " <<radius<< " is erased! Now, we have " <<numOfCircles<< " circles.\n" ;}

    static int getNumOfCircles(){ return numOfCircles;}

     Point &getCenter(){ return center;}

     bool pointInCircle(Point &q)

     {

         int b=(q.getX()-center.getX())*(q.getX()-center.getX())+(q.getY()-center.getY())*(q.getY()-center.getY());

         if (b<radius*radius)

             return 1;

         return 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;

}

转载于:https://www.cnblogs.com/TogetherLaugh/p/6544727.html