python计算平面形体有周长和面积
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2024-03-12 20:10:20
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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())