简化输出有理数 python3

Simplify output rationals python3

我编写了一个程序,它读取 2 个分数和一个运算符,并在评估时给出答案。不要介意代码的长度,我只是添加它是为了完整。我的问题如下:当我作为输入输入时

12/23
23/12
*

我希望它给我输出 1。但是它给了我 1/1。我该如何纠正?

x = input().split('/')
y = input().split('/')
z = input()
def gcd ( a, b ):
    if  b == 0:
        return a
    else:
        return gcd(b, a%b)

class Rational:
    def __init__ ( self, a=0, b=1 ):
        g  =  gcd ( a, b )
        self.n  =  a / g
        self.d  =  b / g
    def __add__ ( self, other ):
        return Rational ( self.n * other.d + other.n * self.d,
                          self.d * other.d )
    def __sub__ ( self, other ):
        return Rational ( self.n * other.d - other.n * self.d,
                          self.d * other.d )
    def __mul__ ( self, other ):
        return  Rational ( self.n * other.n, self.d * other.d )
    def __div__ ( self, other ):
        return  Rational ( self.n * other.d, self.d * other.n )
    def __str__ ( self ):
        return "%d/%d" % ( self.n, self.d )
    def __float__ ( self ):
        return  float ( self.n ) / float ( self.d )

q = Rational()
w = Rational()
q.n = int(x[0])
q.d = int(x[1])
w.n = int(y[0])
w.d = int(y[1])
answer = eval("q"+z+"w")

因为在内部如何存储这两个相等的数字并不重要,所以您唯一需要修改的方法是 __str__,它执行外部表示:

def __str__ ( self ):
    if self.d == 1:
        return "%d" % self.n
    return "%d/%d" % ( self.n, self.d )

这将正确处理 n / 1 的所有情况,包括 1 / 1