如何在Python中遍历十进制数的元素

这并没有那么简单,但就我所能测试的而言,它似乎是有效的。

一个错误修复应该删除分数中多余的0。

它变得越来越混乱,但我必须假设会出现某些输入错误,并将其过滤掉。比如进入1。或1或1.1000,例如,

def find_Odd_Even(m):
    try:
        n = list(str( m ))
        if n.count(".") < 2:
            if n.count(".") == 0:
                n.extend([ ".", "0" ])
            elif n.index(".") == len(n) - 1:
                n.append("0")
            b = n.index(".")
            BB = "".join( n[ b+1: ] )
            while len(BB) > 1:
                k = BB.removesuffix("0")
                if k == BB:
                    break
                else:
                    BB = k
            AA = int("".join( n[ :b ] ))
            BB = int( BB )
            if AA&1 == 1 or BB&1 == 1:
                answer = f"{m} is an odd number"
            else:
                answer = f"{m} is an even number"
        else:
            answer = "Bad input"
    except Exception as err:
        answer = str(err)
    finally:
        return answer

while 1:
    m = input("Enter an decimal number: ")
    if m == "":
        break
    else:
        print(find_Odd_Even( m ))