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);
}