[Codewars] 100: Last digit of a huge number

题目

For a given list [x1, x2, x3, ..., xn] compute the last (decimal) digit of x1 ^ (x2 ^ (x3 ^ (... ^ xn))).

E. g.,

last_digit([3, 4, 2]) == 1

because 3 ^ (4 ^ 2) = 3 ^ 16 = 43046721.

Beware: powers grow incredibly fast. For example, 9 ^ (9 ^ 9) has more than 369 millions of digits. lastDigit has to deal with such numbers efficiently.

Corner cases: we assume that 0 ^ 0 = 1 and that lastDigit of an empty list equals to 1.

This kata generalizes Last digit of a large number; you may find useful to solve it beforehand.

我的答案

def last_digit(lst):
    res = 1
    for i in reversed(lst):
        res = i ** (res if res < 4 else res % 4 + 4)
    return res % 10

其他精彩答案

def last_digit(lst):
    if not lst:
        return 1
    else:
        out = 1
        for n in lst[len(lst):0:-1]:
            out = n**out
            if out > 2:
                out -= 2
                out %= 4
                out += 2
    return lst[0]**out% 10

import functools

def powmod(e, b):
    if e==0 or b==1:
        return 1
    if e==1 or b==0:
        return b
    else:
        return pow(b, e, 20)+20
        

def last_digit(lst):
    return functools.reduce(powmod, lst[::-1], 1) % 10