leetcode-problemset/leetcode-cn/originData/apply-operations-to-maximize-score.json

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        "question": {
            "questionId": "3001",
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            "title": "Apply Operations to Maximize Score",
            "titleSlug": "apply-operations-to-maximize-score",
            "content": "<p>You are given an array <code>nums</code> of <code>n</code> positive integers and an integer <code>k</code>.</p>\n\n<p>Initially, you start with a score of <code>1</code>. You have to maximize your score by applying the following operation at most <code>k</code> times:</p>\n\n<ul>\n\t<li>Choose any <strong>non-empty</strong> subarray <code>nums[l, ..., r]</code> that you haven&#39;t chosen previously.</li>\n\t<li>Choose an element <code>x</code> of <code>nums[l, ..., r]</code> with the highest <strong>prime score</strong>. If multiple such elements exist, choose the one with the smallest index.</li>\n\t<li>Multiply your score by <code>x</code>.</li>\n</ul>\n\n<p>Here, <code>nums[l, ..., r]</code> denotes the subarray of <code>nums</code> starting at index <code>l</code> and ending at the index <code>r</code>, both ends being inclusive.</p>\n\n<p>The <strong>prime score</strong> of an integer <code>x</code> is equal to the number of distinct prime factors of <code>x</code>. For example, the prime score of <code>300</code> is <code>3</code> since <code>300 = 2 * 2 * 3 * 5 * 5</code>.</p>\n\n<p>Return <em>the <strong>maximum possible score</strong> after applying at most </em><code>k</code><em> operations</em>.</p>\n\n<p>Since the answer may be large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [8,3,9,3,8], k = 2\n<strong>Output:</strong> 81\n<strong>Explanation:</strong> 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.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [19,12,14,6,10,18], k = 3\n<strong>Output:</strong> 4788\n<strong>Explanation:</strong> 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</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length == n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= min(n * (n + 1) / 2, 10<sup>9</sup>)</code></li>\n</ul>\n",
            "translatedTitle": "操作使得分最大",
            "translatedContent": "<p>给你一个长度为 <code>n</code>&nbsp;的正整数数组&nbsp;<code>nums</code>&nbsp;和一个整数 <code>k</code>&nbsp;。</p>\n\n<p>一开始，你的分数为 <code>1</code>&nbsp;。你可以进行以下操作至多 <code>k</code>&nbsp;次，目标是使你的分数最大：</p>\n\n<ul>\n\t<li>选择一个之前没有选过的 <strong>非空</strong>&nbsp;子数组&nbsp;<code>nums[l, ..., r]</code>&nbsp;。</li>\n\t<li>从&nbsp;<code>nums[l, ..., r]</code>&nbsp;里面选择一个 <strong>质数分数</strong>&nbsp;最高的元素 <code>x</code>&nbsp;。如果多个元素质数分数相同且最高，选择下标最小的一个。</li>\n\t<li>将你的分数乘以&nbsp;<code>x</code>&nbsp;。</li>\n</ul>\n\n<p><code>nums[l, ..., r]</code>&nbsp;表示&nbsp;<code>nums</code>&nbsp;中起始下标为&nbsp;<code>l</code>&nbsp;，结束下标为 <code>r</code>&nbsp;的子数组，两个端点都包含。</p>\n\n<p>一个整数的 <strong>质数分数</strong>&nbsp;等于&nbsp;<code>x</code>&nbsp;不同质因子的数目。比方说，&nbsp;<code>300</code>&nbsp;的质数分数为&nbsp;<code>3</code>&nbsp;，因为&nbsp;<code>300 = 2 * 2 * 3 * 5 * 5</code>&nbsp;。</p>\n\n<p>请你返回进行至多 <code>k</code>&nbsp;次操作后，可以得到的 <strong>最大分数</strong>&nbsp;。</p>\n\n<p>由于答案可能很大，请你将结果对&nbsp;<code>10<sup>9 </sup>+ 7</code>&nbsp;取余后返回。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [8,3,9,3,8], k = 2\n<b>输出：</b>81\n<b>解释：</b>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。</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [19,12,14,6,10,18], k = 3\n<b>输出：</b>4788\n<b>解释：</b>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</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length == n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= min(n * (n + 1) / 2, 10<sup>9</sup>)</code></li>\n</ul>\n",
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                "<div class=\"_1l1MA\">Calculate <code>nums[i]</code>'s prime score <code>s[i]</code> by factoring in <code>O(sqrt(nums[i]))</code> time.</div>",
                "<div class=\"_1l1MA\">For each <code>nums[i]</code>, find the nearest index <code>left[i]</code> on the left (if any) such that <code>s[left[i]] >= s[i]</code>. if none is found, set <code>left[i]</code> to <code>-1</code>. Similarly, find the nearest index <code>right[i]</code> on the right (if any) such that <code>s[right[i]] > s[i]</code>. If none is found, set <code>right[i]</code> to <code>n</code>.</div>",
                "<div class=\"_1l1MA\">Use a monotonic stack to compute <code>right[i]</code> and <code>left[i]</code>.</div>",
                "<div class=\"_1l1MA\">For each index <code>i</code>, if <code>left[i] + 1 <= l <= i <= r <= right[i] - 1</code>, then <code>s[i]</code> is the maximum value in the range <code>[l, r]</code>. For this particular <code>i</code>, there are <code>ranges[i] = (i - left[i]) * (right[i] - i)</code> ranges where index <code>i</code> will be chosen.</div>",
                "<div class=\"_1l1MA\">Loop over all elements of <code>nums</code> by non-increasing prime score, each element will be chosen <code>min(ranges[i], remainingK)</code> times, where <code>reaminingK</code> denotes the number of remaining operations. Therefore, the score will be multiplied by <code>s[i]^min(ranges[i],remainingK)</code>.</div>",
                "<div class=\"_1l1MA\">Use fast exponentiation to quickly calculate <code>A^B mod C</code>.</div>"
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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