1. Roman numerals 将阿拉伯数字转成罗马数字，可以这样写，不过不是很高效:
def checkio(num): a_num = range(1, 11) + [50, 100, 500, 1000] r_num = ('I', 'II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX', 'X', 'L', 'C', 'D', 'M') num_dict = (dict(zip(a_num, r_num))) res = '' if num == 0: return '' if num in a_num: return num_dict[num] else: str_num = str(num) length = len(str_num) if length == 2: first_num = int(str_num[0]) last_num = int(str_num[1]) if first_num <= 3: res += num_dict[first_num].replace('I', 'X') + num_dict[last_num] elif first_num == 4: res += ('XL' + num_dict[last_num]) elif first_num == 5: res += ('L' + num_dict[last_num]) elif first_num != 9: res += ('L' + num_dict[first_num - 5].replace('I', 'X') + num_dict[last_num]) else: res += ('XC' + num_dict[last_num]) if length == 3: first_num = int(str_num[0]) if first_num <= 3: res += num_dict[first_num].replace('I', 'C') + checkio(int(str_num[1:])) elif first_num == 4: res += ('CD' + checkio(int(str_num[1:]))) elif first_num == 5: res += ('D' + checkio(int(str_num[1:]))) elif first_num != 9: res += ('D' + num_dict[first_num - 5].replace('I', 'C') + checkio(int(str_num[1:]))) else: res += ('CM' + checkio(int(str_num[1:]))) if length == 4: first_num = int(str_num[0]) res += ('M' * first_num + checkio(int(str_num[1:]))) return res
2. Transposed Matrix 转置矩阵 Python CookBook里的方法:
def checkio(data): return map(list, zip(*data))
3. The Angles of a Triangle 给出三角形三条边，求出来三个角度，需要用余弦定理(CosA = (b*b + c*c - a*a)/(2*b*c))
import math def checkio(a, b, c): res = [0, 0, 0] if not (a+b>c and a+c>b and b+c>a): return res else: cos_A = (b**2 + c**2 - a**2) / (2.0*b*c) num1 = int(round(math.degrees(math.acos(cos_A)))) # 四舍五入方法 int(round(num)) cos_B = (c**2 + a**2 - b**2) / (2.0*a*c) num2 = int(round(math.degrees(math.acos(cos_B)))) num3 = 180 - num1 - num2 res = sorted([num1, num2, num3]) return res
4.restricted-sum 计算一个数字列表的加和，但是不许使用sum, for, while, import, reduce这几个keywords,可以写这么一行:
def checkio(alist): return eval('+'.join(map(str,alist)))
如果再加一个不能使用'+'的条件，那就这样:
def checkio(alist): return eval(chr(43).join(map(str,alist)))
5.x-o-referee 判断井字棋游戏是否获胜的问题:
#!/usr/bin/env python #-*- coding:utf-8 -*- def checkio(game_status): # 判断3行 row = [i[0] for i in game_status if i[0] in 'XO' and i.count(i[0]) == 3] if row: return row[0] # 根据中心点判断 center = game_status[1][1] if center in 'XO': if [center, game_status[0][2], game_status[2][0]].count(center) == 3: return center elif [center, game_status[0][0], game_status[2][2]].count(center) == 3: return center # 判断3列 for i in xrange(0, 3): first = game_status[0][i] if first in 'XO': second = game_status[1][i] third = game_status[2][i] if [first, second, third].count(first) == 3: return first return 'D'
6.even-last 把一个list的偶数下标和最后一个元素乘积返回:
def checkio(array = None): if array: #print array[::2] return array[-1] * sum(array[::2]) return 0
7. most-numbers 计算一个list中的最大值和最小值的差:
def checkio(*array): if len(array) == 0: return 0 return round(max(array) - min(array), 3)
8. digits-multiplication 把一个数字的非0数相乘起来：
def checkio(num): return reduce(lambda a,b: a*b, map(lambda s_num: int(s_num) if (s_num!='0') else 1, str(num)))
9.common-words 得到两个字符串按照','分割的相同单词:
def checkio(stra, strb): return ','.join(sorted(list(reduce(lambda a,b: a&b, map(lambda s: set(s), [i.split(',') for i in [stra, strb]])))))
10. fizz-buzz问题:
def checkio(num): if num % 3 == 0 and num % 5 == 0: return "Fizz Buzz" if num % 3 == 0: return "Fizz" if num % 5 == 0: return "Buzz" return str(num)