Description
Given a positive integer K, you need find the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.
Return the length of N. If there is no such N, return -1.
给定正整数 K，你需要找出可以被 K 整除的、仅包含数字 1 的最小正整数 N。
返回 N 的长度。如果不存在这样的 N，就返回 -1。
题目链接：https://leetcode.com/problems/smallest-integer-divisible-by-k/
Difficulty: medium
Example 1:
Input: 1
Output: 1
Explanation: The smallest answer is N = 1, which has length 1.
Example 2:
Input: 2
Output: -1
Explanation: There is no such positive integer N divisible by 2.
Example 3:
Input: 3
Output: 3
Explanation: The smallest answer is N = 111, which has length 3.
Note:
- 1 <= K <= 10^5
分析
- 已知任意数字的个位数为li=[1,3,7,9]中任意一个，其任意倍数的个位数不可能为1；
- 若K的个位数在li中，返回-1；
- 反之，N表示n位长的数字且每一位都为“1”，判断是否可以整除K；
- 若整除，返回N的长度，反之，长度加一，回到上一步。
参考代码
class Solution(object):
def smallestRepunitDivByK(self, K):
li=[1,3,7,9]
if(K%10 not in li):
return -1
index=1
N=1
while(N%K != 0):
index+=1
N=N*10+1
return index