Question

我正在编写一个程序，对输入的int进行各种处理，但是如果其中一位数字为9，希望它执行不同的操作。我已经四处搜寻，但找不到适合的方法，到目前为止，这是我的代码：

Num = int(input("Enter a large number: "))
if Num  # does not contain 9
   Num2 = Num + 10
   print("If I added 10 to your number it would be:", Num2) 
   print("The new number times 2 would be", Num2 * 2)
   count = 0
   number = Num
   if (number.valueof(x).contains("9")):
       print("No 9s plz")
   else: 
       while (number > 9):
           number = number // 10
           count = count + 1
           for i in range(1, count + 1):
               Num = Num + 1 * (10**i)
               i = i+1
           print("adding one to every digit on your first number would be:", Num + 1)
else:
   print("Sorry we dont do 9s")

Answer 1

希望这会有所帮助。

Num = int(input("Enter a large number: "))
if Num!=9: #does not contain 9
    Num2 = Num+10 
    print("If I added 10 to your number it would be:" , Num2) 
    print("The new number times 2 would be " , Num2*2) 
    count = 0; number = Num; 
    set1=set(str(number))
    if "9" in set1: 
        print("No 9s plz") 
    else: 
        while (number > 9): 
            number = number//10 
            count = count + 1 
            for i in range(1,count+1): 
                Num = Num+1*(10**i) 
                i = i+1 
                print("adding one to every digit on your first number would be: " , Num + 1)
else:
    print("Sorry we dont do 9s")

如何检查数字是否包含特定数字

1 个答案: