标签：blog os io for 代码 div log c
罗马数字的题目, 注意几个关键的数字即可: (100, 400, 500, 900) -> (‘C‘, ‘CD‘, ‘D‘, ‘CM‘); (10, 40, 50, 90)->(‘X‘, ‘XL‘, ‘L‘, ‘XC‘)
1 def checkio(data):
2 rel = ‘‘
3
4 thonsand = data / 1000
5 rel += thonsand * ‘M‘
6
7 data %= 1000
8
9 table = [[‘C‘, ‘CD‘, ‘D‘, ‘CM‘], [‘X‘, ‘XL‘, ‘L‘, ‘XC‘]]
10
11 pos = 100
12
13 for i in range(0, 2):
14 bit = data / pos
15 if bit > 0:
16 if bit < 4:
17 rel += bit * table[i][0]
18 elif bit == 4:
19 rel += table[i][1]
20 elif bit == 5:
21 rel += table[i][2]
22 elif bit == 9:
23 rel += table[i][3]
24 else:
25 rel += (table[i][2] + table[i][0] * (bit - 5))
26
27 data %= pos
28 pos /= 10
29
30 if data > 0:
31 unit = [‘I‘, ‘II‘, ‘III‘, ‘IV‘, ‘V‘, ‘VI‘, ‘VII‘, ‘VIII‘, ‘IX‘, ‘X‘]
32 rel += unit[data - 1]
33
34 #replace this for solution
35 return rel
另外还需注意没有个位数的情况, 即第30行所示
观摩JulianNicholls的代码
1 elements = { 1000 : ‘M‘, 900 : ‘CM‘, 500 : ‘D‘, 400 : ‘CD‘,
2 100 : ‘C‘, 90 : ‘XC‘, 50 : ‘L‘, 40: ‘XL‘,
3 10 : ‘X‘, 9 : ‘IX‘, 5 : ‘V‘, 4: ‘IV‘, 1 : ‘I‘ }
4
5 def checkio(data):
6 roman = ‘‘
7
8 for n in sorted(elements.keys(), reverse=True):
9 while data >= n:
10 roman += elements[n]
11 data -= n
12
13 return roman
sorted(elements.keys(), reverse=True), 按key从大到小排列; 思路不错, 比数位对应的值大, 即加上该数位对应的字母, 并减去对应的数值, 直到data为0
标签：blog os io for 代码 div log c
原文地址：http://www.cnblogs.com/hzhesi/p/3891669.html
