python计算平面形体有周长和面积

4、对平面形体有周长和面积,周长、面积应怎样计算(用什么方法)?要求实现运行时的多态性。请编程,并测试。

Shape

正方形(Square) 长方形(Rectangle) 圆形(Circle) 圆环(Annulus)

#
import math


class Shape:

    def __init__(self, w, length=0) -> None:
        super().__init__()
        self.w = w
        self.length = length

    def area(self):
        pass

    def perimeter(self):
        pass


class Square(Shape):

    def __init__(self, w, length=0) -> None:
        super().__init__(w, length)

    def area(self):
        s = self.w ** 2
        return s

    def perimeter(self):
        c = 4 * self.w
        return c


# s1 = Square(4)
# print(s1.perimeter())
# print(s1.area())
class Rectangle(Shape):

    def __init__(self, w, length) -> None:
        super().__init__(w, length)

    def area(self):
        s = self.w * self.length
        return s

    def perimeter(self):
        c = (self.w + self.length) * 2
        return c


# r1 = Rectangle(4, 5)
# print(r1.perimeter())
# print(r1.area())
class Circle(Shape):

    def __init__(self, w, length=0) -> None:
        super().__init__(w, length)

    def area(self):
        s = math.pi * self.w * self.w
        return s

    def perimeter(self):
        c = math.pi * self.w * 2
        return c


# c1 = Circle(4)
# print(c1.perimeter())
# print(c1.area())
class Annulus(Shape):

    def __init__(self, w, length) -> None:
        super().__init__(w, length)

    def area(self):
        s = abs(math.pi * self.w * self.w - math.pi * self.length * self.length)
        return s

    def perimeter(self):
        c = math.pi * self.w * 2 + math.pi * self.length * 2
        return c


# a1 = Annulus(4, 5)
# print(a1.perimeter())
# print(a1.area())

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转载自blog.csdn.net/qq_40679091/article/details/109716042