---

CheckiO 是面向初學(xué)者和高級程序員的編碼游戲，使用 Python 和 JavaScript 解決棘手的挑戰(zhàn)和有趣的任務(wù)，從而提高你的編碼技能，本博客主要記錄自己用 Python 在闖關(guān)時的做題思路和實(shí)現(xiàn)代碼，同時也學(xué)習(xí)學(xué)習(xí)其他大神寫的代碼。

CheckiO 官網(wǎng)：https://checkio.org/

我的 CheckiO 主頁：https://py.checkio.org/user/TRHX/

CheckiO 題解系列專欄：https://itrhx.blog.csdn.net/category_9536424.html

CheckiO 所有題解源代碼：https://github.com/TRHX/Python-CheckiO-Exercise

---

題目描述

【Roman Numerals】：將阿拉伯?dāng)?shù)字轉(zhuǎn)換成羅馬數(shù)字

【鏈接】：https://py.checkio.org/mission/roman-numerals/

【輸入】：一個整數(shù)形式的阿拉伯?dāng)?shù)字

【輸出】：一個字符串形式的羅馬數(shù)字

【前提】：0 < number < 4000

【范例】：

checkio(6) == 'VI' checkio(76) == 'LXXVI' checkio(13) == 'XIII' checkio(44) == 'XLIV' checkio(3999) == 'MMMCMXCIX'

解題思路

先將一到十、十位整數(shù)和百位整數(shù)以列表形式列出來，分為四種情況：一位數(shù)、兩位數(shù)、三位數(shù)和四位數(shù)，每種情況還要判斷是否為整數(shù)，不是整數(shù)的，將其除以位數(shù)的余，再次傳給 checkio() 函數(shù)，求剩下的值。

這種方法比較繁瑣，語法簡單，不適合大一點(diǎn)的數(shù)字，太菜了，看看大神的解答吧

代碼實(shí)現(xiàn)

def checkio(data):roman_list1 = ['I', 'II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX']roman_list2 = ['X', 'XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC']roman_list3 = ['C', 'CC', 'CCC', 'CD', 'D', 'DC', 'DCC', 'DCCC', 'CM']if 0 < data < 10:roman = roman_list1[data - 1]return romanelif 10 <= data < 100:if data % 10 == 0:roman = roman_list2[data // 10 - 1]else:roman = roman_list2[data // 10 - 1] + checkio(data % 10)return romanelif 100 <= data < 1000:if data % 100 == 0:roman = roman_list3[data // 100 - 1]else:roman = roman_list3[data // 100 - 1] + checkio(data % 100)return romanelif 1000 <= data < 9999:if data % 1000 == 0:roman = 'M' * (data // 1000)else:roman = 'M' * (data // 1000) + checkio(data % 1000)return romanif __name__ == '__main__':# These "asserts" using only for self-checking and not necessary for auto-testingassert checkio(6) == 'VI', '6'assert checkio(76) == 'LXXVI', '76'assert checkio(499) == 'CDXCIX', '499'assert checkio(3888) == 'MMMDCCCLXXXVIII', '3888'print('Done! Go Check!')

大神解答

大神解答 NO.1

def checkio(n):result = ''for arabic, roman in zip((1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1),'M CM D CD C XC L XL X IX V IV I'.split()):result += n // arabic * romann %= arabicreturn result

大神解答 NO.2

roman1 = ('', 'M', 'MM', 'MMM') roman2 = ('', 'C', 'CC', 'CCC', 'CD', 'D', 'DC', 'DCC', 'DCCC', 'CM') roman3 = ('', 'X', 'XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC') roman4 = ('', 'I', 'II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX')def checkio(data):data1 = int(data / 1000)data2 = int(data % 1000 / 100)data3 = int(data % 100 / 10)data4 = int(data % 10 )return roman1[data1] + roman2[data2] + roman3[data3] + roman4[data4]

大神解答 NO.3

def checkio(data):s = ''ones = ['X','I', 'II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX']tens = ['C', 'X', 'XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC']mils = ['M', 'C', 'CC', 'CCC', 'CD', 'D', 'DC', 'DCC', 'DCCC', 'CM']if data / 1000 != 0:s = s + 'M'*(data/1000)data = data % 1000if data / 100 != 0:s = s + mils[data/100]data = data % 100if data / 10 != 0:s = s + tens[data/10]data = data % 10if data / 1 != 0:s = s + ones[data/1]return s

大神解答 NO.4

from enum import Enumclass Roman(Enum):M = 1000CM = 900D = 500CD = 400C = 100XC = 90L = 50XL = 40X = 10IX = 9V = 5IV = 4I = 1@classmethoddef encode(cls, n):for numeral in cls:rep, n = divmod(n, numeral.value)yield numeral.name * repcheckio = lambda n: ''.join(Roman.encode(n))

大神解答 NO.5

def checkio(data):base = "I"*database = base.replace("I"*5, "V")base = base.replace("V"*2, "X")base = base.replace("X"*5, "L")base = base.replace("L"*2, "C")base = base.replace("C"*5, "D")base = base.replace("D"*2, "M")base = base.replace("DCCCC", "CM")base = base.replace("CCCC", "CD")base = base.replace("LXXXX", "XC")base = base.replace("XXXX", "XL")base = base.replace("VIIII", "IX")base = base.replace("IIII", "IV")return base 創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎勵來咯，堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎

總結(jié)

以上是生活随笔為你收集整理的【Python CheckiO 题解】Roman Numerals的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。