#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;

class shape
{
public:
virtual double area()const{return 0.0;}
virtual double zhouchang()const{return 0.0;}
};

class sanjiaoxing :public shape
{double a,b,c;
public:
sanjiaoxing(double x,double y,double z){a=x;b=y,c=z;}

virtual double area()const
    {  double p;
       p=(a+b+c)/2;
       cout<<"三角形的面积是 "<<endl;
      return(sqrt(p*(p-a)*(p-b)*(p-c)));
     }
virtual double zhouchang()const{cout<<"三角形的周长是 "<<endl;return(a+b+c);}

 

void showa(){cout<<"三角形的第一条边 "<<a<<endl;}
void showb(){cout<<"三角形的第二条边 "<<b<<endl;}
void showc(){cout<<"三角形的第三条边 "<<c<<endl;}

 

bool panduan(){if((a+b>c)&&(abs(a-b)<c)&&(a+c>b)&&(abs(a-c)<b&&(b+c>a)&&(abs(b-c)<a))) return 1;else return 0;}

};

 

class circle:public shape
   {   double r;
       public:
       circle(double m){r=m;}
       virtual double area()const{cout<<"圆的面积是 "<<endl;return(3.14*r*r);}
       virtual double zhouchang()const{cout<<"圆的的周长是 "<<endl;return(2*3.14*r);}
       void showr(){cout<<"圆的半径"<<r<<endl;}

  };

 

class juxing:public shape
{ double chang,kuan;
  public:
  juxing(double x,double y){chang=x;kuan=y;}
  virtual double area()const{cout<<"矩形的面积是"<<endl;return(chang*kuan);}
    virtual double zhouchang()const{cout<<"矩形周长是 "<<endl;return(chang*2+kuan*2);}
    void showchang(){cout<<"矩形的长为 "<<chang<<endl;}
 void showkuan(){cout<<"矩形的宽为"<<kuan<<endl;}

};

 

 

 

void fn(const shape& s)
{cout<<s.area()<<endl;
cout<<s.zhouchang ()<<endl;

}

 

 

 

 

void main()
{
 
 
 
 circle m(3);
    m. showr(); 
 fn(m);
   

 

juxing n(3,5);
n. showchang();
n. showkuan();
fn(n);

sanjiaoxing p(3,4,7);
p. showa();
p. showb();
p. showc();
p.panduan ();
fn(p);

}

判断三条边能不能构成三角形 有点问题  帮我改改吧

 

 

 

就是如果能构成三角形的话  执行输出周长和面积  如果构不成的话就不执行