Python基于递归算法求最小公倍数和最大公约数示例_Python

Python基于递归算法求最小公倍数和最大公约数示例

2021-03-23 00:17混沌鳄鱼 Python

这篇文章主要介绍了Python基于递归算法求最小公倍数和最大公约数,结合实例形式分析了Python使用递归算法进行数值计算的相关操作技巧,需要的朋友可以参考下

本文实例讲述了Python基于递归算法求最小公倍数和最大公约数。分享给大家供大家参考，具体如下：

									# 最小公倍数

									def lcm(a, b, c=1):

									  if a * c % b != 0:

									    return lcm(a, b, c+1)

									  else:

									    return a*c

									test_cases = [(4, 8), (35, 42), (5, 7), (20, 10)]

									for case in test_cases:

									  print('lcm of {} & {} is {}'.format(*case, lcm(*case)))

									def lcm(a, b):

									  for i in range(2, min(a,b)+1):

									    if a % i == 0 and b % i == 0:

									      return i * lcm(a//i, b//i)

									  else:

									    return a*b

									test_cases = [(4, 8), (5, 7), (24, 16), (35, 42)]

									for case in test_cases:

									  print('lcm of {} & {} is {}'.format(*case, lcm(*case)))

									# 最大公约数

									def gcd(a, b):

									  if a == b:

									    return a

									  elif a-b > b:

									    return gcd(a-b, b)

									  else:

									    return gcd(b, a-b)

									test_cases = [(35, 14), (88, 66), (5, 4), (20, 10)]

									for case in test_cases:

									  print('GCD of {} & {} is {}'.format(*case, gcd(*case)))

运行结果：

lcm of 4 & 8 is 8
lcm of 35 & 42 is 210
lcm of 5 & 7 is 35
lcm of 20 & 10 is 20
GCD of 35 & 14 is 7
GCD of 88 & 66 is 22
GCD of 5 & 4 is 1
GCD of 20 & 10 is 10

希望本文所述对大家Python程序设计有所帮助。

原文链接：https://blog.csdn.net/xpresslink/article/details/78875473