零一数字

Question

更新：输入的数字> 44超时。关于如何阻止其超时的任何想法？ 我的代码应该输出最小的数字，该数字仅包含输入分为的1或0。在此示例中，仅包含1或0的34的最小倍数是111010。但是，我的代码仅输出无限循环...有什么想法吗？

#example input: 34
#example ouput: 111010
#2<n<1000
print("Please enter your number...")
n = int(input())
counter = 0

mylist = ['2','3','4','5','6','7','8','9']

t = n
check = str(t)
while any(x in check for x in mylist):
    counter += 1
    t = n * counter
    continue
else:
    print(t)

Answer 1

简单地将给定数字加到一个和上，直到它的字符串仅包含0和1：

# you already solved the input part, so skipping it here
n = 34
summed = n
allowed = {"0","1"}
while set(str(summed)) - allowed:   #  @MadPhysicists suggestion instead of 
                                          #  any(x not in allowed for x in str(summed)):
    summed += n
else:
    print(summed)

输出：

111010

不知道，但是从长远来看，我猜是加法比乘法快。