簡化輸出合理性python3

Question

我寫了一個程序，讀取2個分數和一個運算符，並在評估時給出答案。不要介意代碼的長度，我只是添加它是完整的。我的問題是這樣的：當我輸入

12/23 
23/12 
* 

我希望它給我的輸出1進入。但它給了我1/1。我該如何解決這個問題？

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") 

Answer 1

由於這不要緊，你如何存儲兩個人數相等的內部，你需要修改的唯一方法是__str__，這確實外在表現：

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

這將處理的所有案件n/1正確，包括1/1。