leetcode-problemset/leetcode-cn/originData/sum-of-squares-of-special-elements.json

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            "title": "Sum of Squares of Special Elements ",
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            "content": "<p>You are given a <strong>1-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>\n\n<p>An element <code>nums[i]</code> of <code>nums</code> is called <strong>special</strong> if <code>i</code> divides <code>n</code>, i.e. <code>n % i == 0</code>.</p>\n\n<p>Return <em>the <strong>sum of the squares</strong> of all <strong>special</strong> elements of </em><code>nums</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4]\n<strong>Output:</strong> 21\n<strong>Explanation:</strong> There are exactly 3 special elements in nums: nums[1] since 1 divides 4, nums[2] since 2 divides 4, and nums[4] since 4 divides 4. \nHence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21.  \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,7,1,19,18,3]\n<strong>Output:</strong> 63\n<strong>Explanation:</strong> There are exactly 4 special elements in nums: nums[1] since 1 divides 6, nums[2] since 2 divides 6, nums[3] since 3 divides 6, and nums[6] since 6 divides 6. \nHence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63. \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;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n</ul>\n",
            "translatedTitle": "特殊元素平方和",
            "translatedContent": "<p>给你一个下标从 <strong>1</strong> 开始、长度为 <code>n</code> 的整数数组 <code>nums</code> 。</p>\n\n<p>对 <code>nums</code> 中的元素 <code>nums[i]</code> 而言，如果 <code>n</code> 能够被 <code>i</code> 整除，即 <code>n % i == 0</code> ，则认为 <code>num[i]</code> 是一个 <strong>特殊元素</strong> 。</p>\n\n<p>返回 <code>nums</code> 中所有 <strong>特殊元素</strong> 的 <strong>平方和</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3,4]\n<strong>输出：</strong>21\n<strong>解释：</strong>nums 中共有 3 个特殊元素：nums[1]，因为 4 被 1 整除；nums[2]，因为 4 被 2 整除；以及 nums[4]，因为 4 被 4 整除。 \n因此，nums 中所有特殊元素的平方和等于 nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21 。  \n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,7,1,19,18,3]\n<strong>输出：</strong>63\n<strong>解释：</strong>nums 中共有 4 个特殊元素：nums[1]，因为 6 被 1 整除；nums[2] ，因为 6 被 2 整除；nums[3]，因为 6 被 3 整除；以及 nums[6]，因为 6 被 6 整除。 \n因此，nums 中所有特殊元素的平方和等于 nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63 。 </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;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n</ul>\n",
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