1015. Smallest Integer Divisible by K (M)
Smallest Integer Divisible by K (M)
题目
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 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
题意
找出一个最长的全部由1组成的整数N,使其能被K整除。
思路
由于N的长度不定,不能直接用普通遍历去做。记由n个1组成的整数为\(f(n)\),而\(f(n)\)除以K的余数为\(g(n)\),则有\(g(n+1)=(g(n)*10+1)\%k\),下证:
\[\begin{cases}
f(n)\div K=a \\
g(n)=f(n)\ \%\ K \\
f(n+1)=f(n)\times10+1 \\
g(n+1)=f(n+1)\ \%\ K
\end{cases}
\ \Rightarrow\
\begin{cases}
f(n)=a\times K\ +g(n) \\
g(n+1)=(f(n)\times10+1)\ \%\ K
\end{cases}
\ \Rightarrow\
g(n+1)=(10a \times K + 10 \times g(n)+1)\ \%\ K \equiv (10\times g(n)+1)\ \%\ K
\]
所以可以每次都用余数去处理。
另一个问题是确定循环的次数。对于除数K,得到的余数最多有0~K-1这K种情况,因此当我们循环K次都没有找到整除时,其中一定有重复的余数,这意味着之后的循环也不可能整除。所以最多循环K-1次。
代码实现
Java
class Solution {
public int smallestRepunitDivByK(int K) {
if (K % 5 == 0 || K % 2 == 0) {
return -1;
}
int len = 1, n = 1;
for (int i = 0; i < K; i++) {
if (n % K == 0) {
return len;
}
len++;
n = (n * 10 + 1) % K;
}
return -1;
}
}
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