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：
[egin{cases}
f(n)div K=a \
g(n)=f(n) \% K \
f(n+1)=f(n) imes10+1 \
g(n+1)=f(n+1) \% K
end{cases}
Rightarrow
egin{cases}
f(n)=a imes K +g(n) \
g(n+1)=(f(n) imes10+1) \% K
end{cases}
Rightarrow
g(n+1)=(10a imes K + 10 imes g(n)+1) \% K equiv (10 imes 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;
}
}