1015 Smallest Integer Divisible by K 可被 K 整除的最小整数
Description:
Given a positive integer k, you need to find the length of 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.
Note:
n may not fit in a 64-bit signed integer.
Example:
Example 1:
Input: k = 1
Output: 1
Explanation: The smallest answer is n = 1, which has length 1.
Example 2:
Input: k = 2
Output: -1
Explanation: There is no such positive integer n divisible by 2.
Example 3:
Input: k = 3
Output: 3
Explanation: The smallest answer is n = 111, which has length 3.
Constraints:
1 <= k <= 10^5
题目描述:
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 的长度。
返回 n 的长度。如果不存在这样的 n ，就返回-1。
注意：
n 不符合 64 位带符号整数。
示例 :
示例 1：
输入：k = 1
输出：1
解释：最小的答案是 n = 1，其长度为 1。
示例 2：
输入：k = 2
输出：-1
解释：不存在可被 2 整除的正整数 n 。
示例 3：
输入：k = 3
输出：3
解释：最小的答案是 n = 111，其长度为 3。
提示:
1 <= k <= 10^5
思路:
数学模拟
cur 从 1 开始寻找, 下一个 next = cur * 10 + 1
因为 cur % k != 0, 令 cur = p * k + q, 其中 p, q 为整数
由于 next % k = (cur * 10 + 1) % k = ((pk + q) * 10 + 1) % k = (10pk + 10q + 1) % k
又由于 10pk % k 恒等于 0
所以 next % k = (10q + 1) % k, 注意到 cur = pk + q, 所以 cur % k = (pk + q) % k = q
所以 next % k = (10q + 1) % k = (10 * cur) % k + 1
时间复杂度为 O(n), 空间复杂度为 O(1)
代码:
C++:
class Solution
{
public:
int smallestRepunitDivByK(int k)
{
if (!(k & 1) or !(k % 5)) return -1;
int cur = 1, result = 1;
while (cur % k)
{
++result;
cur = (cur * 10) % k + 1;
}
return result;
}
};
Java:
class Solution {
public int smallestRepunitDivByK(int k) {
if ((k & 1) == 0 || k % 5 == 0) return -1;
int cur = 1, result = 1;
while (cur % k != 0) {
++result;
cur = (cur * 10) % k + 1;
}
return result;
}
}
Python:
class Solution:
def smallestRepunitDivByK(self, k: int) -> int:
if str(k % 10) not in "1379":
return -1
num = "1" * len(str(k))
result, remainder = len(num), int(num) % k
while remainder:
remainder = (10 * remainder + 1) % k
result += 1
return result