小于 X 且可被 Y 整除的最大自然数

最佳解决方案

5000 // 39 * 39

#4992

解决方案比较

def test1():
    num = max(i for i in reversed(range(5000)) if i % 39 == 0)


def test2():
    num = max(filter(lambda x: x % 39 == 0, reversed(range(5000))))


def test3():
    for i in range(5000, 39, -1):
      if not i % 39:
        break


def test4():
    5000 // 39 * 39


if __name__ == '__main__':
    import timeit
    setup="from __main__ import test1,test2,test3, test4"


    print('test1:',  timeit.timeit('test1()', setup=setup,number=10000))
    print('-' * 25)
    print('test2:',  timeit.timeit('test2()', setup=setup,number=10000))
    print('-' * 25)
    print('test3:',  timeit.timeit('test3()', setup=setup,number=10000))
    print('-' * 25)
    print('test4:',  timeit.timeit('test4()', setup=setup,number=10000))

#test1: 3.1136840023100376
#-------------------------
#test2: 6.391721844673157
#-------------------------
#test3: 0.008747044950723648
#-------------------------
#test4: 0.0008490197360515594

结论：您可以更快地解决 - 不使用循环，仅基于数学运算。