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Simple solution in Clear category for Roman Numerals by cheikhnamouna
def checkio(num):
s = ""
for k, v in [(1000, ["", "M", "MM", "MMM"]),
(100, ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM", "M"]),
(10, ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"]),
(1, ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"])]:
s += v[num // k]
num %= k
return s
Dec. 31, 2018