首页 > 代码库 > Problem E: 点在圆内吗?
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;
}
Problem E: 点在圆内吗?
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。