<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the number of <strong>subarrays</strong> of </em><code>nums</code><em> where the least common multiple of the subarray&#39;s elements is </em><code>k</code>.</p>

<p>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</p>

<p>The <strong>least common multiple of an array</strong> is the smallest positive integer that is divisible by all the array elements.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,6,2,7,1], k = 6
<strong>Output:</strong> 4
<strong>Explanation:</strong> The subarrays of nums where 6 is the least common multiple of all the subarray&#39;s elements are:
- [<u><strong>3</strong></u>,<u><strong>6</strong></u>,2,7,1]
- [<u><strong>3</strong></u>,<u><strong>6</strong></u>,<u><strong>2</strong></u>,7,1]
- [3,<u><strong>6</strong></u>,2,7,1]
- [3,<u><strong>6</strong></u>,<u><strong>2</strong></u>,7,1]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3], k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no subarrays of nums where 2 is the least common multiple of all the subarray&#39;s elements.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i], k &lt;= 1000</code></li>
</ul>

class Solution:
    def subarrayLCM(self, nums: List[int], k: int) -> int:
        k_dividable = []
        k_dividable_len = [0] * len(nums)
        for num in nums:
            k_dividable.append(k % num == 0)
        for i, dv in enumerate(reversed(k_dividable)):
            if dv:
                if i > 0:
                    k_dividable_len[-1 - i] = k_dividable_len[-1 - (i - 1)] + 1
                else:
                    k_dividable_len[-1 - i] = 1
        ans = 0
        # print(k_dividable_len)
        for i, num in enumerate(nums):
            base = num
            idx = i
            while idx < len(nums) and k % base == 0:
                if k == base:
                    # print(i, idx, k_dividable_len[idx])
                    ans += k_dividable_len[idx]
                    break
                idx += 1
                if idx < len(nums):
                    base = math.lcm(base, nums[idx])
        return ans
107