Problem D: 点在圆内吗?Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 4448 Solved: 2258
[Submit][Status]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:
- 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。
- 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。 - 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。
- 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.
HINTAppend Codeappend.cc,
#include <iostream>
using namespace std;
class Point
{
friend class Circle;
private:
int x;
int 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(){numOfPoints--;cout<<"A Point ("<<x<<", "<<y<<") is erased! 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;}
};
int Point::numOfPoints=0;
class Circle
{
friend class Point;
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(){numOfCircles--;cout<<"A circle at ("<<center.x<<", "<<center.y<<") and radius "<<radius<<" is erased! 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 copied! Now, we have "<<numOfCircles<<" circles."<<endl;}
static int getNumOfCircles(){return numOfCircles;}
Point &getCenter(){return center;}
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 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;
}