📅  最后修改于: 2023-12-03 14:54:38.108000             🧑  作者: Mango
有一块圆形蛋糕需要被切成8个等份,但是只能使用3次切割。请写出一个算法,完成这个任务。
我们可以使用递归和分治的思想,将蛋糕分成若干个小块,再分别切割,最后合并成8个等份。
具体步骤如下:
def cut_cake(cake, cuts):
if cuts == 0:
return [cake]
pieces = []
if cuts == 1:
# 切成4个小块
r = cake.radius
d = r * 2
c1 = Circle(cake.x - r, cake.y, r)
c2 = Circle(cake.x + r, cake.y, r)
c3 = Circle(cake.x, cake.y - r, r)
c4 = Circle(cake.x, cake.y + r, r)
pieces.extend([c1, c2, c3, c4])
else:
# 切成4个小块并递归切割
sub_cakes = cut_cake(cake, cuts - 1)
for sub_cake in sub_cakes:
sub_pieces = cut_cake(sub_cake, cuts - 1)
pieces.extend(sub_pieces)
# 拼接小块
result = []
for i, piece in enumerate(pieces):
if piece not in result:
for j, other_piece in enumerate(pieces):
if i != j and other_piece not in result:
if are_adjacent(piece, other_piece):
result.append(join_pieces(piece, other_piece))
break
else:
result.append(piece)
return result
class Circle(object):
def __init__(self, x, y, r):
self.x = x
self.y = y
self.radius = r
def are_adjacent(c1, c2):
return ((c1.x - c2.x) ** 2 + (c1.y - c2.y) ** 2) == (c1.radius + c2.radius) ** 2
def join_pieces(c1, c2):
assert are_adjacent(c1, c2)
c1_center = (c1.x, c1.y)
c2_center = (c2.x, c2.y)
# 计算缺口的位置
x, y = midpoint(c1_center, c2_center)
dx = x - c1.x
dy = y - c1.y
missing_point = (c1.x + dx, c1.y + dy)
# 计算拼接后的圆心与半径
new_center = midpoint(c1_center, c2_center)
new_radius = int(c1.radius * sin(pi / 4))
return Circle(*new_center, new_radius)
def midpoint(p1, p2):
return ((p1[0] + p2[0]) // 2, (p1[1] + p2[1]) // 2)
if __name__ == '__main__':
cake = Circle(0, 0, 16)
pieces = cut_cake(cake, 3)
for i, piece in enumerate(pieces):
print(f'Piece {i + 1}: ({piece.x}, {piece.y}), {piece.radius}')
本文介绍了如何用递归和分治的思想,使用3次切割将圆形蛋糕切成8个等份。代码实现了切割和拼接的功能,可以用于解决类似的问题。