对于类与对象的简单运用（代码）

// Note:Your choice is C++ IDE
#include <iostream>
#include <cmath>
using namespace std;
class Point //直角座标系的Point(点)类
{

public:
double x; //将x,y设为公用变量，以便派生类函数可以访问
double y;
Point(){x=0;y=0;} //默认构造函数
Point(double X,double Y){x=X;y=Y;} //实现初始化的构造函数
double Distance(Point,Point); //类中的成员函数
void midPoint(Point,Point,Point);
void slope(Point,Point);
};

double Point::Distance(Point c1,Point c2) //成员函数Distance计算两点间距离
{
double l;
l=sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y)); //计算公式
return (l);
}

void Point::midPoint(Point c1,Point c2,Point c3) //成员函数midpoint计算两点的中点坐标
{
c3.x=(c1.x+c2.x)/2;
c3.y=(c1.y+c2.y)/2;
cout<<"两点的中点坐标是:("<<c3.x<<","<<c3.y<<")"<<endl;
}

void Point::slope(Point c1,Point c2) //成员函数slope计算两点的斜率
{
double n;
n=(c2.y-c1.y)/(c2.x-c1.x);
cout<<"两点的斜率:"<<n<<endl;
}

class Circle:private Point //以Point类为基类，派生出一个Circle(圆)类
{ public:
double r; //新增数据变量半径r
public:
Circle(double X,double Y,double R):Point(X,Y) {r=R;}
void circumference(Circle);
void area(Circle);
void check(double,double,Circle);
void judge(double,Circle,Circle);
};

void Circle::circumference(Circle R1) //成员函数circumference计算圆的周长
{ double C,p=3.141;
C=2*p*R1.r;
cout<<"圆的周长:"<<C<<endl;
}

void Circle::area(Circle R1) //成员函数area计算圆的面积
{
double S,p=3.141;
S=R1.r*R1.r;
cout<<"圆的面积:"<<S<<endl;
}

void Circle::check(double m,double L,Circle R1) //成员函数check判断点在圆内、圆上、圆外
{
if(m==R1.r)
{cout<<"所以c1(1,1)在以（0，0）为圆心且以2为半径的圆R1上"<<endl;} //用if语句进行判断
if(m<R1.r)
{cout<<"所以c1(1,1)在以（0，0）为圆心且以2为半径的圆R1内"<<endl;}
if(m>R1.r)
{cout<<"所以c1(1,1)在以（0，0）为圆心且以2为半径的圆R1外"<<endl;}
if(L==R1.r)
{cout<<"所以c2(0,2)在以（0，0）为圆心且以2为半径的圆R1上"<<endl;}
if(L<R1.r)
{cout<<"所以c2(0,2)在以（0，0）为圆心且以2为半径的圆R1内"<<endl;}
if(L>R1.r)
{cout<<"所以c2(0,2)在以（0，0）为圆心且以2为半径的圆R1外"<<endl;}
}
void Circle::judge(double t,Circle R1,Circle R2) //判断两圆的位置关系
{
cout<<"两圆R1与R2的半径分别为"<<R1.r<<","<<R2.r<<endl;
if(R1.r<R2.r)R1.r=R2.r;
if(t==(R1.r+R2.r))
{cout<<"所以圆R1与圆R2外切"<<endl;}
if(t==(R1.r-R2.r))
{cout<<"所以圆R1与圆R2内切"<<endl;}
if(t<(R1.r+R2.r)&&t>(R1.r-R2.r))
{cout<<"所以圆R1与圆R2相交"<<endl;}
if(t>(R1.r+R2.r))
{cout<<"所以圆R1与圆R2外离"<<endl;}
if(t<(R1.r-R2.r)&&t>0)
{cout<<"所以圆R1与圆R2内含"<<endl;}
if(t==0)
{cout<<"所以圆R1与圆R2同心"<<endl;}

}
class Line:private Point //以Point类为基类，派生出一个Line(线)类
{private:
double k;
double b;
public:
//构造函数
Line(double k0,double b0):k(k0),b(b0){}
Line(double x0,double y0,double k0):Point(x0,y0),k(k0){}
Line(double x1,double y1,double k0,double b0):Point(x1,y1),k(k0),b(b0){}

void judge_valuea(Line,Line);
void judge_valueb(Line,Point);
double Dis(Line,Point);
void judge_valuec(Line,double,Point,Circle);
void display_l();
void line_q(Point,Point,Circle);
};
void Line::display_l()
{
if(b>0)cout<<"直线 Y="<<k<<"*X+"<<b;
else if(b==0) cout<<"直线 Y="<<k<<"*X";
else cout<<"直线 Y="<<k<<"*X"<<b;
}

void Line::judge_valuea( Line L1,Line L2) //判断两直线关系
{
if(L1.k==L2.k)cout<<"两线平行"<<endl;
else cout<<"两线相交"<<endl;
cout<<"两线相交的交点为"<<"("<<(L1.b-L2.b)/(L2.k-L1.k)<<"," //计算两线的交点坐标
<<((L1.b-L2.b)/(L2.k-L1.k))*L1.k+L1.b<<")"<<endl;
}
void Line::judge_valueb(Line L4,Point c1) //判断点与直线的关系
{
if(c1.y==L4.k*c1.x+L4.b)cout<<"点c1（1，1）在线y="<<L4.k<<"*x+"<<L4.b<<"上"<<endl;
else cout<<"点c1（1，1）不在线y="<<L4.k<<"*x+"<<L4.b<<"上"<<endl;
}
double Line::Dis(Line L3,Point c5) //计算点到直线的距离
{
return(fabs((L3.k*c5.x-c5.y+L3.b)/(sqrt(1.0+L3.k*L3.k)))); //点到直线的距离公式
}

void Line::judge_valuec(Line ,double i,Point c5,Circle R2) //判断线与圆的关系
{
double a,b1,c;
long double x1,x2,y1,y2;

a=1+k*k; //计算线与圆的交点，通过圆与线方程联立求得
b1=2*k*(b-c5.y)-2*c5.x;
c=(b-c5.y)*(b-c5.y)+c5.x*c5.x-R2.r*R2.r;
x1=((-b1)+sqrt(b1*b1-4*a*c))/2*a;
y1=k*x1+b;
x2=((-b1)-sqrt(b1*b1-4*a*c))/2*a;
y2=k*x2+b;
if(i==2)
{
cout<<"所以线与圆的位置关系为相切"<<endl;
cout<<"线与圆的切点为("<<x1<<","<<y2<<")"<<endl;
}
if(i<2)
{
cout<<"所以线与圆的位置关系为相交"<<endl;
cout<<"线与圆的交点为:"<<"交点一("<<x1<<","<<y1<<");交点二("<<x2<<","<<y2<<")"<<endl;

}
if(i>2)cout<<"所以线与圆的位置关系为相离"<<endl;

}

void Line::line_q(Point c5,Point c6,Circle R2)
{
double s,t;
double g;
double k1=(c6.y-c5.y)/(c6.x-c5.x);
double m,n;
s=sqrt((c6.x-c5.x)*(c6.x-c5.x)+(c6.y-c5.y)*(c6.y-c5.y));
t=sqrt((s*s-R2.r*R2.r));
g=R2.r/t;
m=(g+k1)/(1-g*k1);
n=(g+m)/(1-g*m);
double b1=c5.y-m*c5.x;
double b2=c5.y-n*c5.x;
Line L9(m,b1);
Line L_1(n,b2);
cout<<"与圆相切的直线分别为：";
L9.display_l();
cout<<" "<<endl;
L_1.display_l();
}

class Triangle:private Point //以Point类为基类，派生出一个Triangle（三角形）类
{private:
double a;
double b;
double c;

public:
Triangle(){a=0;b=0;c=0;}
Triangle(double a0,double b0,double c0){a=a0;b=b0;c=c0;}
Triangle(double x3,double y3,double x4,double y4,double x5,double y5)
{Point c3(x3,y3);
Point c4(x4,y4);
Point c5(x5,y5);
a=c3.Distance(c3,c4);
b=c3.Distance(c3,c5);
c=c4.Distance(c4,c5);
}
void judge_valued();
void judge_valuee(Triangle,Point,Point,Point);
};
void Triangle::judge_valued()
{
if(a+b>c&&a+c>b&&b+c>a)
{
cout<<"此三边能够成三角形！"<<endl<<"三角形的面积为：";
double l=(a+b+c)/2;
double s=sqrt(l*(l-a)*(l-b)*(l-c));
cout<<s<<endl;
}
else
{
cout<<"输入错误，不能构成三角形故不能进行计算！"<<endl;
}
}
void Triangle::judge_valuee(Triangle T1,Point c5,Point c7,Point c8)
{
Point c1=c7;
Point c2=c8;

double a=c1.Distance(c1,c2);
double b=c1.Distance(c1,c5);
double c=c2.Distance(c2,c5);
Triangle t(a,b,c);
cout<<"由直线与圆的交点坐标和圆心所构成的";
t.judge_valued();
}

int main() //主函数
{ double j,m,L,t,i;
Point c1(1,1),c2(0,2),c3,c4(0,0),c5(2,3),c6(5,6),c7(1,2),c8(2,6);
Circle R1(0,0,2),R2(2,3,2); //设置圆R1和圆R2的值
Line L1(0,0,3,-1),L2(0,0,2,2);
Line L4(0,0,1,1);
Line L3(0,0,1,-1);
Line L5(0,0,0,1);
j=c1.Distance(c1,c2); //通过对象c1调用另一个对象c2进行运算执行函数Distance
cout<<"两点的距离是："<<j<<endl;
c3.midPoint(c1,c2,c3);
c1.slope(c1,c2);
R1.circumference(R1);
R1.area(R1);
m=c1.Distance(c1,c4);
cout<<"点c1(1,1)到圆心(0,0)的距离是："<<m<<endl;
L=c2.Distance(c2,c4);
cout<<"点c2(0,2)到圆心(2,3)的距离是："<<L<<endl;
R1.check(m,L,R1);
t=c4.Distance(c4,c5);
cout<<"两圆心的距离为："<<t<<endl; //计算两圆的圆心的距离
R1.judge(t,R1,R2); //通过对象R1进行判断
cout<<"线L1:y=3x-1与线L2:y=2x+2的关系为：";
L1.judge_valuea(L1,L2);
L4.judge_valueb(L4,c1);
i=L3.Dis(L3,c5); //计算圆心到直线的距离
cout<<"以2为半径的圆R2的圆心（2，3）到直线L3:y=x-1的距离为："<<i<<endl;
L3.judge_valuec(L3,i,c5,R2);
i=L5.Dis(L5,c5);
cout<<"以2为半径的圆R2的圆心（2，3）到直线L5:y=1的距离为："<<i<<endl;
L5.judge_valuec(L5,i,c5,R2);
Triangle T1(2,3,1,2,2,6);
T1.judge_valuee(T1,c5,c7,c8);
L3.line_q(c5,c6,R2);
return 0;
}