mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
192 lines
24 KiB
JSON
192 lines
24 KiB
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"content": "<p>In LeetCode Store, there are <code>n</code> items to sell. Each item has a price. However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.</p>\n\n<p>You are given an integer array <code>price</code> where <code>price[i]</code> is the price of the <code>i<sup>th</sup></code> item, and an integer array <code>needs</code> where <code>needs[i]</code> is the number of pieces of the <code>i<sup>th</sup></code> item you want to buy.</p>\n\n<p>You are also given an array <code>special</code> where <code>special[i]</code> is of size <code>n + 1</code> where <code>special[i][j]</code> is the number of pieces of the <code>j<sup>th</sup></code> item in the <code>i<sup>th</sup></code> offer and <code>special[i][n]</code> (i.e., the last integer in the array) is the price of the <code>i<sup>th</sup></code> offer.</p>\n\n<p>Return <em>the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers</em>. You are not allowed to buy more items than you want, even if that would lower the overall price. You could use any of the special offers as many times as you want.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> price = [2,5], special = [[3,0,5],[1,2,10]], needs = [3,2]\n<strong>Output:</strong> 14\n<strong>Explanation:</strong> There are two kinds of items, A and B. Their prices are $2 and $5 respectively. \nIn special offer 1, you can pay $5 for 3A and 0B\nIn special offer 2, you can pay $10 for 1A and 2B. \nYou need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> price = [2,3,4], special = [[1,1,0,4],[2,2,1,9]], needs = [1,2,1]\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> The price of A is $2, and $3 for B, $4 for C. \nYou may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. \nYou need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. \nYou cannot add more items, though only $9 for 2A ,2B and 1C.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == price.length</code></li>\n\t<li><code>n == needs.length</code></li>\n\t<li><code>1 <= n <= 6</code></li>\n\t<li><code>0 <= price[i] <= 10</code></li>\n\t<li><code>0 <= needs[i] <= 10</code></li>\n\t<li><code>1 <= special.length <= 100</code></li>\n\t<li><code>special[i].length == n + 1</code></li>\n\t<li><code>0 <= special[i][j] <= 50</code></li>\n</ul>\n",
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"translatedTitle": "大礼包",
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"translatedContent": "<p>在 LeetCode 商店中， 有 <code>n</code> 件在售的物品。每件物品都有对应的价格。然而，也有一些大礼包，每个大礼包以优惠的价格捆绑销售一组物品。</p>\n\n<p>给你一个整数数组 <code>price</code> 表示物品价格，其中 <code>price[i]</code> 是第 <code>i</code> 件物品的价格。另有一个整数数组 <code>needs</code> 表示购物清单，其中 <code>needs[i]</code> 是需要购买第 <code>i</code> 件物品的数量。</p>\n\n<p>还有一个数组 <code>special</code> 表示大礼包，<code>special[i]</code> 的长度为 <code>n + 1</code> ，其中 <code>special[i][j]</code> 表示第 <code>i</code> 个大礼包中内含第 <code>j</code> 件物品的数量，且 <code>special[i][n]</code> （也就是数组中的最后一个整数）为第 <code>i</code> 个大礼包的价格。</p>\n\n<p>返回<strong> 确切 </strong>满足购物清单所需花费的最低价格，你可以充分利用大礼包的优惠活动。你不能购买超出购物清单指定数量的物品，即使那样会降低整体价格。任意大礼包可无限次购买。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>price = [2,5], special = [[3,0,5],[1,2,10]], needs = [3,2]\n<strong>输出：</strong>14\n<strong>解释：</strong>有 A 和 B 两种物品，价格分别为 ¥2 和 ¥5 。 \n大礼包 1 ，你可以以 ¥5 的价格购买 3A 和 0B 。 \n大礼包 2 ，你可以以 ¥10 的价格购买 1A 和 2B 。 \n需要购买 3 个 A 和 2 个 B ， 所以付 ¥10 购买 1A 和 2B（大礼包 2），以及 ¥4 购买 2A 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>price = [2,3,4], special = [[1,1,0,4],[2,2,1,9]], needs = [1,2,1]\n<strong>输出：</strong>11\n<strong>解释：</strong>A ，B ，C 的价格分别为 ¥2 ，¥3 ，¥4 。\n可以用 ¥4 购买 1A 和 1B ，也可以用 ¥9 购买 2A ，2B 和 1C 。 \n需要买 1A ，2B 和 1C ，所以付 ¥4 买 1A 和 1B（大礼包 1），以及 ¥3 购买 1B ， ¥4 购买 1C 。 \n不可以购买超出待购清单的物品，尽管购买大礼包 2 更加便宜。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == price.length</code></li>\n\t<li><code>n == needs.length</code></li>\n\t<li><code>1 <= n <= 6</code></li>\n\t<li><code>0 <= price[i] <= 10</code></li>\n\t<li><code>0 <= needs[i] <= 10</code></li>\n\t<li><code>1 <= special.length <= 100</code></li>\n\t<li><code>special[i].length == n + 1</code></li>\n\t<li><code>0 <= special[i][j] <= 50</code></li>\n</ul>\n",
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