2019-2020-1 second week of the semester 20,192,428 jobs - Arabic digital conversion Roman numerals

Python Programming on Roman numerals converted to Arabic numerals

Arabic notation as a traditional method of counting position, wants to be a Roman numeral conversion is a difficult thing, the law of the Roman numeral I summarized as follows:

1. Same digital ligatures, equal to the number indicated by the numbers obtained by adding the number ( at most three ligatures )
2. Small numbers of large numbers on the right side, is equal to the number indicated by the number obtained by adding the numbers
3. A low number ( limited Ⅰ, X and C ) the number of large numbers on the left, equal to the number represented by the reduced number of large numbers obtained

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Therefore, with the len function , conventional loop and judgment statement, the basis of print, input and other functions, I successfully edited out of this program.

Code is as follows :( can learn, plagiarism is strictly prohibited )

#阿拉伯数字转换罗马数字1~3999
#I=1 V=5 X=10 L=50 C=100 D=500 M=1000 IV=4 IX=9 XL=40 XC=90 CD=400 CM=900
#在构成数字的时候，有下列规则：
#1、相同的数字连写，所表示的数等于这些数字相加得到的数
#2、小的数字在大的数字的右边，所表示的数等于这些数字相加得到的数
#3、小的数字，（限于Ⅰ、X和C）在大的数字的左边，所表示的数等于大数减小数得到的数

print('''--------------------------------------------------
作者：20192428魏来
阿拉伯数字向罗马数字的转换
--------------------------------------------------''')
list = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}
a = 0
while a < 9999 : #多次进行转换:（9998次）
    a = a+1
    romannum = input('''--------------------------------------------------
请输入罗马数字(输入范围：1-3999)：''')
    result = 0
    omit = -1  # omit为需省略的字符位
    if romannum in list:
        print(list[romannum])
    else:
        for b in range(len(romannum)):  # 遍历字符串的每一位:(0,len-1)
            if b == omit:  # 如果无需省略b
                pass
            elif b <= len(romannum)-2:  # 如果还没有遍历到最后一位:(len-1-1)
                if list[romannum[b]] < list[romannum[b+1]]:  # 如果前一位对应值小于后一位的
                    result += list[romannum[b+1]] - list[romannum[b]]  # 进行减操作
                    omit = b+1  # 省略下一位字符
                else:
                    result += list[romannum[b]]
            else:  #  遍历到最后一位
                result += list[romannum[b]]
        print('''对应阿拉伯数字为：{}
--------------------------------------------------'''.format(result))

If additions or improvements welcome message.

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