You are given an array nums of n positive integers and an integer k.
Initially, you start with a score of 1. You have to maximize your score by applying the following operation at most k times:
nums[l, ..., r] that you haven't chosen previously.x of nums[l, ..., r] with the highest prime score. If multiple such elements exist, choose the one with the smallest index.x.Here, nums[l, ..., r] denotes the subarray of nums starting at index l and ending at the index r, both ends being inclusive.
The prime score of an integer x is equal to the number of distinct prime factors of x. For example, the prime score of 300 is 3 since 300 = 2 * 2 * 3 * 5 * 5.
Return the maximum possible score after applying at most k operations.
Since the answer may be large, return it modulo 109 + 7.
\n
Example 1:
\n\n\nInput: nums = [8,3,9,3,8], k = 2\nOutput: 81\nExplanation: To get a score of 81, we can apply the following operations:\n- Choose subarray nums[2, ..., 2]. nums[2] is the only element in this subarray. Hence, we multiply the score by nums[2]. The score becomes 1 * 9 = 9.\n- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 1, but nums[2] has the smaller index. Hence, we multiply the score by nums[2]. The score becomes 9 * 9 = 81.\nIt can be proven that 81 is the highest score one can obtain.\n\n
Example 2:
\n\n\nInput: nums = [19,12,14,6,10,18], k = 3\nOutput: 4788\nExplanation: To get a score of 4788, we can apply the following operations: \n- Choose subarray nums[0, ..., 0]. nums[0] is the only element in this subarray. Hence, we multiply the score by nums[0]. The score becomes 1 * 19 = 19.\n- Choose subarray nums[5, ..., 5]. nums[5] is the only element in this subarray. Hence, we multiply the score by nums[5]. The score becomes 19 * 18 = 342.\n- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 2, but nums[2] has the smaller index. Hence, we multipy the score by nums[2]. The score becomes 342 * 14 = 4788.\nIt can be proven that 4788 is the highest score one can obtain.\n\n\n
\n
Constraints:
\n\n1 <= nums.length == n <= 1051 <= nums[i] <= 1051 <= k <= min(n * (n + 1) / 2, 109)给你一个长度为 n 的正整数数组 nums 和一个整数 k 。
一开始，你的分数为 1 。你可以进行以下操作至多 k 次，目标是使你的分数最大：
nums[l, ..., r] 。nums[l, ..., r] 里面选择一个 质数分数 最高的元素 x 。如果多个元素质数分数相同且最高，选择下标最小的一个。x 。nums[l, ..., r] 表示 nums 中起始下标为 l ，结束下标为 r 的子数组，两个端点都包含。
一个整数的 质数分数 等于 x 不同质因子的数目。比方说， 300 的质数分数为 3 ，因为 300 = 2 * 2 * 3 * 5 * 5 。
请你返回进行至多 k 次操作后，可以得到的 最大分数 。
由于答案可能很大，请你将结果对 109 + 7 取余后返回。
\n\n
示例 1：
\n\n\n输入：nums = [8,3,9,3,8], k = 2\n输出：81\n解释：进行以下操作可以得到分数 81 ：\n- 选择子数组 nums[2, ..., 2] 。nums[2] 是子数组中唯一的元素。所以我们将分数乘以 nums[2] ，分数变为 1 * 9 = 9 。\n- 选择子数组 nums[2, ..., 3] 。nums[2] 和 nums[3] 质数分数都为 1 ，但是 nums[2] 下标更小。所以我们将分数乘以 nums[2] ，分数变为 9 * 9 = 81 。\n81 是可以得到的最高得分。\n\n
示例 2：
\n\n\n输入：nums = [19,12,14,6,10,18], k = 3\n输出：4788\n解释：进行以下操作可以得到分数 4788 ：\n- 选择子数组 nums[0, ..., 0] 。nums[0] 是子数组中唯一的元素。所以我们将分数乘以 nums[0] ，分数变为 1 * 19 = 19 。\n- 选择子数组 nums[5, ..., 5] 。nums[5] 是子数组中唯一的元素。所以我们将分数乘以 nums[5] ，分数变为 19 * 18 = 342 。\n- 选择子数组 nums[2, ..., 3] 。nums[2] 和 nums[3] 质数分数都为 2，但是 nums[2] 下标更小。所以我们将分数乘以 nums[2] ，分数变为 342 * 14 = 4788 。\n4788 是可以得到的最高的分。\n\n\n
\n\n
提示：
\n\n1 <= nums.length == n <= 1051 <= nums[i] <= 1051 <= k <= min(n * (n + 1) / 2, 109)nums[i]'s prime score s[i] by factoring in O(sqrt(nums[i])) time.nums[i], find the nearest index left[i] on the left (if any) such that s[left[i]] >= s[i]. if none is found, set left[i] to -1. Similarly, find the nearest index right[i] on the right (if any) such that s[right[i]] > s[i]. If none is found, set right[i] to n.right[i] and left[i].i, if left[i] + 1 <= l <= i <= r <= right[i] - 1, then s[i] is the maximum value in the range [l, r]. For this particular i, there are ranges[i] = (i - left[i]) * (right[i] - i) ranges where index i will be chosen.nums by non-increasing prime score, each element will be chosen min(ranges[i], remainingK) times, where reaminingK denotes the number of remaining operations. Therefore, the score will be multiplied by s[i]^min(ranges[i],remainingK).A^B mod C.\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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