4.25

问题描述：

              定义抽象基类Shape，由它派生出五个派生类：Circle（圆形）、Square（正方形）、Rectangle（ 长方形）、Trapezoid （梯形）和Triangle （三角形），用虚函数分别计算各种图形的面积，并求出它们的和。要求用基类指针数组。使它的每一个元素指向一个派生类的对象。PI=3.1415926

输入格式：

      例如：输入在一行中给出9个大于0的数，用空格分隔，分别代表圆的半径，正方形的边长，矩形的宽和高，梯形的上底、下底和高，三角形的底和高。

输出格式：

      例如：输出所有图形的面积和，小数点后保留3位有效数字。

 

代码示例：

#include<iostream>
using namespace std;
#define PI 3.14159
class Shape
{
protected:
    double a, b, c;

public:

    Shape(double a=0,double b=0,double c=0)
    {
        this->a = a; this->b = b;
        this->c = c;
    }

    virtual double getmianji() = 0;

};

class Circle:public Shape{
public:
    Circle(double a) :Shape(a) {}
    double getmianji(){
        return PI * a * a;
    }

};

class Square :public Shape
{
public:
    Square(double a) :Shape(a) {}
    double getmianji(){
        return  a * a;
    }
};

class Rectangle :public Shape{
public:
    Rectangle(double a,double b) :Shape(a,b) {}
    double getmianji()    {
        return a*b;
    }
};

class Trapezoid :public Shape{
public:
    Trapezoid(double a,double b,double c) :Shape(a,b,c) {}
    double getmianji(){
        return (a + b) * c * 0.5;
    }
};
class Triangle :public Shape//（三角形）
{
public:
    Triangle(double a,double b) :Shape(a,b) {}
    double getmianji(){
      return a*b*0.5;
    }
};

int main(){
    double a, b, c;
    cin >> a;
    Circle C(a);
    cin >> a ;
    Square s(a);
    cin >> a >> b;
    Rectangle r(a, b);
    cin >> a >> b >> c;
    Trapezoid t1(a, b, c);
    cin >> a >> b;
    Triangle t2(a, b);
    Shape* sp[5] = { &C,&s,&r,&t1,&t2 };
    double sum = 0;
    for (int i = 0; i < 5; i++)
    {
        sum += sp[i]->getmianji();
    }
    printf("total of all areas = %0.3f", sum);
}