leetcode-problemset/leetcode-cn/originData/collecting-chocolates.json

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            "title": "Collecting Chocolates",
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            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index <code>i</code>&nbsp;is <code>nums[i]</code>. Each chocolate is of a different type, and initially, the chocolate at the index&nbsp;<code>i</code>&nbsp;is of <code>i<sup>th</sup></code> type.</p>\n\n<p>In one operation, you can do the following with an incurred <strong>cost</strong> of <code>x</code>:</p>\n\n<ul>\n\t<li>Simultaneously change the chocolate of <code>i<sup>th</sup></code> type to <code>((i + 1) mod n)<sup>th</sup></code> type for all chocolates.</li>\n</ul>\n\n<p>Return <em>the minimum cost to collect chocolates of all types, given that you can perform as many operations as you would like.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [20,1,15], x = 5\n<strong>Output:</strong> 13\n<strong>Explanation:</strong> Initially, the chocolate types are [0,1,2]. We will buy the 1<sup>st</sup>&nbsp;type of chocolate at a cost of 1.\nNow, we will perform the operation at a cost of 5, and the types of chocolates will become [1,2,0]. We will buy the 2<sup>nd</sup><sup> </sup>type of chocolate at a cost of 1.\nNow, we will again perform the operation at a cost of 5, and the chocolate types will become [2,0,1]. We will buy the 0<sup>th </sup>type of chocolate at a cost of 1. \nThus, the total cost will become (1 + 5 + 1 + 5 + 1) = 13. We can prove that this is optimal.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3], x = 4\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> We will collect all three types of chocolates at their own price without performing any operations. Therefore, the total cost is 1 + 2 + 3 = 6.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= x &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "收集巧克力",
            "translatedContent": "<p>给你一个长度为 <code>n</code> 、下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，表示收集不同巧克力的成本。每个巧克力都对应一个不同的类型，最初，位于下标 <code>i</code> 的巧克力就对应第 <code>i</code> 个类型。</p>\n\n<p>在一步操作中，你可以用成本 <code>x</code> 执行下述行为：</p>\n\n<ul>\n\t<li>同时修改所有巧克力的类型，将巧克力的类型&nbsp;<code>i<sup>th</sup></code>&nbsp;修改为类型&nbsp;<code>((i + 1) mod n)<sup>th</sup></code>。</li>\n</ul>\n\n<p>假设你可以执行任意次操作，请返回收集所有类型巧克力所需的最小成本。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [20,1,15], x = 5\n<strong>输出：</strong>13\n<strong>解释：</strong>最开始，巧克力的类型分别是 [0,1,2] 。我们可以用成本 1 购买第 1 个类型的巧克力。\n接着，我们用成本 5 执行一次操作，巧克力的类型变更为 [1,2,0] 。我们可以用成本 1 购买第 2 个类型的巧克力。\n然后，我们用成本 5 执行一次操作，巧克力的类型变更为 [2,0,1] 。我们可以用成本 1 购买第 0 个类型的巧克力。\n因此，收集所有类型的巧克力需要的总成本是 (1 + 5 + 1 + 5 + 1) = 13 。可以证明这是一种最优方案。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3], x = 4\n<strong>输出：</strong>6\n<strong>解释：</strong>我们将会按最初的成本收集全部三个类型的巧克力，而不需执行任何操作。因此，收集所有类型的巧克力需要的总成本是 1 + 2 + 3 = 6 。\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 &lt;= 1000</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= x &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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