Given an integer array nums and two integers k and p, return the number of distinct subarrays, which have at most k elements that are divisible by p.
Two arrays nums1 and nums2 are said to be distinct if:
i where nums1[i] != nums2[i].A subarray is defined as a non-empty contiguous sequence of elements in an array.
\n\n\n
Example 1:
\n\n\nInput: nums = [2,3,3,2,2], k = 2, p = 2\nOutput: 11\nExplanation:\nThe elements at indices 0, 3, and 4 are divisible by p = 2.\nThe 11 distinct subarrays which have at most k = 2 elements divisible by 2 are:\n[2], [2,3], [2,3,3], [2,3,3,2], [3], [3,3], [3,3,2], [3,3,2,2], [3,2], [3,2,2], and [2,2].\nNote that the subarrays [2] and [3] occur more than once in nums, but they should each be counted only once.\nThe subarray [2,3,3,2,2] should not be counted because it has 3 elements that are divisible by 2.\n\n\n
Example 2:
\n\n\nInput: nums = [1,2,3,4], k = 4, p = 1\nOutput: 10\nExplanation:\nAll element of nums are divisible by p = 1.\nAlso, every subarray of nums will have at most 4 elements that are divisible by 1.\nSince all subarrays are distinct, the total number of subarrays satisfying all the constraints is 10.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 2001 <= nums[i], p <= 2001 <= k <= nums.length\n
Follow up:
\n\nCan you solve this problem in O(n2) time complexity?
\n", "translatedTitle": "含最多 K 个可整除元素的子数组", "translatedContent": "给你一个整数数组 nums 和两个整数 k 和 p ，找出并返回满足要求的不同的子数组数，要求子数组中最多 k 个可被 p 整除的元素。
如果满足下述条件之一，则认为数组 nums1 和 nums2 是 不同 数组：
i 满足 nums1[i] != nums2[i] 。子数组 定义为：数组中的连续元素组成的一个 非空 序列。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [2,3,3,2,2], k = 2, p = 2\n输出：11\n解释：\n位于下标 0、3 和 4 的元素都可以被 p = 2 整除。\n共计 11 个不同子数组都满足最多含 k = 2 个可以被 2 整除的元素：\n[2]、[2,3]、[2,3,3]、[2,3,3,2]、[3]、[3,3]、[3,3,2]、[3,3,2,2]、[3,2]、[3,2,2] 和 [2,2] 。\n注意，尽管子数组 [2] 和 [3] 在 nums 中出现不止一次，但统计时只计数一次。\n子数组 [2,3,3,2,2] 不满足条件，因为其中有 3 个元素可以被 2 整除。\n\n\n
示例 2：
\n\n\n输入：nums = [1,2,3,4], k = 4, p = 1\n输出：10\n解释：\nnums 中的所有元素都可以被 p = 1 整除。\n此外，nums 中的每个子数组都满足最多 4 个元素可以被 1 整除。\n因为所有子数组互不相同，因此满足所有限制条件的子数组总数为 10 。\n\n\n
\n\n
提示：
\n\n1 <= nums.length <= 2001 <= nums[i], p <= 2001 <= k <= nums.length\n\n
进阶：
\n\n你可以设计并实现时间复杂度为 O(n2) 的算法解决此问题吗？
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