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"question": {
"questionId": "1000438",
"questionFrontendId": "LCP 55",
"categoryTitle": "Algorithms",
"boundTopicId": 1438375,
"title": "采集果实",
"titleSlug": "PTXy4P",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "采集果实",
"translatedContent": "欢迎各位勇者来到力扣新手村,本次训练内容为「采集果实」。\n\n在新手村中各位勇者需要采集一些果实来制作药剂。`time[i]` 表示勇者每次采集 `1limit` 颗第 `i` 种类型的果实需要的时间(即每次最多可以采集 `limit` 颗果实)。\n\n当前勇者需要完成「采集若干批果实」的任务 `fruits[j] = [type, num]` 表示第 `j` 批需要采集 `num` 颗 `type` 类型的果实。采集规则如下:\n- 按 `fruits` 给定的顺序**依次**采集每一批次\n- 采集完当前批次的果实才能开始采集下一批次\n- 勇者完成当前批次的采集后将**清空背包**(即多余的果实将清空)\n\n请计算并返回勇者完成采集任务最少需要的时间。\n\n\n**示例 1**\n>输入:`time = [2,3,2], fruits = [[0,2],[1,4],[2,1]], limit = 3`\n>\n>输出:`10`\n>\n>解释:\n>由于单次最多采集 3 颗\n>第 0 批需要采集 2 颗第 0 类型果实,需要采集 1 次,耗时为 2\\*1=2\n>第 1 批需要采集 4 颗第 1 类型果实,需要采集 2 次,耗时为 3\\*2=6\n>第 2 批需要采集 1 颗第 2 类型果实,需要采集 1 次,耗时为 2\\*1=2\n>返回总耗时 2+6+2=10\n\n**示例 2**\n>输入:`time = [1], fruits = [[0,3],[0,5]], limit = 2`\n>\n>输出:`5`\n>\n>解释:\n>由于单次最多采集 2 颗\n>第 0 批需要采集 3 颗第 0 类型果实,需要采集 2 次,耗时为 1\\*2=2\n>第 1 批需要采集 5 颗第 0 类型果实,需要采集 3 次,耗时为 1\\*3=3\n>需按照顺序依次采集,返回 2+3=5\n\n**提示:**\n- `1 <= time.length <= 100`\n- `1 <= time[i] <= 100`\n- `1 <= fruits.length <= 10^3`\n- `0 <= fruits[i][0] < time.length`\n- `1 <= fruits[i][1] < 10^3`\n- `1 <= limit <= 100`",
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"code": "class Solution {\npublic:\n int getMinimumTime(vector<int>& time, vector<vector<int>>& fruits, int limit) {\n\n }\n};",
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