2016-10-11 80 views
-1
import math 
def roundup(x): 
return int(math.ceil(x/10.0)) * 10 
w=0 
while w == 5: 
print("Would you like to *work out* a missing letter in a GTIN-8 code, or *check* a code?") 
response = input(":") 
if response == 'work out': 
    print("Input a 7 digit GTIN-8 code and I'll work out the 8th") 
    c1 = int(input("Enter FIRST number: ")) 
    c2 = int(input("Enter SECOND number: ")) 
    c3 = int(input("Enter THIRD number: ")) 
    c4 = int(input("Enter FOURTH number: ")) 
    c5 = int(input("Enter FIFTH number: ")) 
    c6 = int(input("Enter SIXTH number: ")) 
    c7 = int(input("Enter SEVENTH number: ")) 

    y = (c1*3+c2+c3*3+c4+c5*3+c6+c7*3) 
    ru2=roundup(y) 
    GTIN8 = ru2-y 
    print("Your GTIN8 Code would be: "+str(c1)+str(c2)+str(c3)+str(c4)+str(c5)+str(c6)+str(c7)+str(GTIN8)) 
    print("Wanna work out another?") 
if response == 'check': 
    print("Input a 8 digit GTIN-8 code and I'll check if it's correct") 
c1 = int(input("Enter FIRST number: ")) 
c2 = int(input("Enter SECOND number: ")) 
c3 = int(input("Enter THIRD number: ")) 
c4 = int(input("Enter FOURTH number: ")) 
c5 = int(input("Enter FIFTH number: ")) 
c6 = int(input("Enter SIXTH number: ")) 
c7 = int(input("Enter SEVENTH number: ")) 
c8 = int(input("Enter EIGTH number: ")) 
y = (c1*3+c2+c3*3+c4+c5*3+c6+c7*3) 
ru2=roundup(y) 
GTIN8 = ru2-y 
if GTIN8 != c8: 
    print("Nope that product code is incorrect!") 
    reply=input("Want to know the correct answer to your code? Type yes if so: ") 
if reply == 'yes': 
    print("The correct answer would have been: "+str(GTIN8)) 
if GTIN8 == c8: 
    print("That code is correct!") 

問題是我已經一次又一次地嘗試使這個代碼更小。我如何重構此代碼以縮短它?

即使通過輸入'response'作爲字符串,以允許用戶一次輸入代碼。

如果你還不知道,這是一個GTIN-8產品代碼的代碼,我知道還有其他各種GTIN-8代碼,但我無法做到,更不用說複製了。

+1

您想要改進的功能代碼可以提交給代碼審查stackexchange –

+0

爲輸入您的7個數字創建一個函數,然後在兩種情況下重用該函數。對於需要第8個輸入的那個,或者保留現在的行,或者讓你的函數使用bool來表示是否需要請求第8個輸入,然後可以返回包含所有值的列表。 –

+0

我建議你在http://codereview.stackexchange.com/上提問 –

回答

-1

爲了讓您一開始,這裏是降低線

c = [int(x) for x in input("Input a 8 digit GTIN-8 code and I'll check if it's correct").split("")] 

的數量現在你可以用c[n]訪問每個角色的一個簡單方法。