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WHnhjV.json
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{
"data": {
"question": {
"questionId": "1000433",
"questionFrontendId": "LCP 50",
"categoryTitle": "Algorithms",
"boundTopicId": 1425580,
"title": "宝石补给",
"titleSlug": "WHnhjV",
"content": "",
"translatedTitle": "宝石补给",
"translatedContent": "欢迎各位勇者来到力扣新手村，在开始试炼之前，请各位勇者先进行「宝石补给」。\n\n每位勇者初始都拥有一些能量宝石， `gem[i]` 表示第 `i` 位勇者的宝石数量。现在这些勇者们进行了一系列的赠送，`operations[j] = [x, y]` 表示在第 `j` 次的赠送中 第 `x` 位勇者将自己一半的宝石（需向下取整）赠送给第 `y` 位勇者。\n\n在完成所有的赠送后，请找到拥有**最多**宝石的勇者和拥有**最少**宝石的勇者，并返回他们二者的宝石数量**之差**。\n\n**注意：**\n- 赠送将按顺序逐步进行。\n\n**示例 1：**\n>输入：`gem = [3,1,2], operations = [[0,2],[2,1],[2,0]]`\n>\n>输出：`2`\n>\n>解释：\n>第 1 次操作，勇者 `0` 将一半的宝石赠送给勇者 `2`， `gem = [2,1,3]`\n>第 2 次操作，勇者 `2` 将一半的宝石赠送给勇者 `1`， `gem = [2,2,2]`\n>第 3 次操作，勇者 `2` 将一半的宝石赠送给勇者 `0`， `gem = [3,2,1]`\n>返回 3 - 1 = 2\n\n**示例 2：**\n>输入：`gem = [100,0,50,100], operations = [[0,2],[0,1],[3,0],[3,0]]`\n>\n>输出：`75`\n>\n>解释：\n>第 1 次操作，勇者 `0` 将一半的宝石赠送给勇者 `2`， `gem = [50,0,100,100]`\n>第 2 次操作，勇者 `0` 将一半的宝石赠送给勇者 `1`， `gem = [25,25,100,100]`\n>第 3 次操作，勇者 `3` 将一半的宝石赠送给勇者 `0`， `gem = [75,25,100,50]`\n>第 4 次操作，勇者 `3` 将一半的宝石赠送给勇者 `0`， `gem = [100,25,100,25]`\n>返回 100 - 25 = 75\n\n**示例 3：**\n>输入：`gem = [0,0,0,0], operations = [[1,2],[3,1],[1,2]]`\n>\n>输出：`0`\n\n**提示：**\n- `2 <= gem.length <= 10^3`\n- `0 <= gem[i] <= 10^3`\n- `0 <= operations.length <= 10^4`\n- `operations[i].length == 2`\n- `0 <= operations[i][0], operations[i][1] < gem.length`",
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"code": "class Solution {\npublic:\n int giveGem(vector<int>& gem, vector<vector<int>>& operations) {\n\n }\n};",
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