Python基于递归算法求最小公倍数和最大公约数示例
时间:2022-01-11 10:28 作者:admin
本文实例讲述了python/' target='_blank'>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*ctest_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*btest_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
PS:这里再为大家推荐一款本站相关在线工具供大家参考:
在线最小公倍数/最大公约数计算工具:
http://tools.jb51.net/jisuanqi/gbs_gys_calc
更多关于Python相关内容感兴趣的读者可查看本站专题:《Python数学运算技巧总结》、《Python数据结构与算法教程》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》、《Python入门与进阶经典教程》及《Python文件与目录操作技巧汇总》
希望本文所述对大家Python程序设计有所帮助。
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