mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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176 lines
28 KiB
JSON
176 lines
28 KiB
JSON
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"title": "Maximum Number of Alloys",
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"content": "<p>You are the owner of a company that creates alloys using various types of metals. There are <code>n</code> different types of metals available, and you have access to <code>k</code> machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy.</p>\n\n<p>For the <code>i<sup>th</sup></code> machine to create an alloy, it needs <code>composition[i][j]</code> units of metal of type <code>j</code>. Initially, you have <code>stock[i]</code> units of metal type <code>i</code>, and purchasing one unit of metal type <code>i</code> costs <code>cost[i]</code> coins.</p>\n\n<p>Given integers <code>n</code>, <code>k</code>, <code>budget</code>, a <strong>1-indexed</strong> 2D array <code>composition</code>, and <strong>1-indexed</strong> arrays <code>stock</code> and <code>cost</code>, your goal is to <strong>maximize</strong> the number of alloys the company can create while staying within the budget of <code>budget</code> coins.</p>\n\n<p><strong>All alloys must be created with the same machine.</strong></p>\n\n<p>Return <em>the maximum number of alloys that the company can create</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> It is optimal to use the 1<sup>st</sup> machine to create alloys.\nTo create 2 alloys we need to buy the:\n- 2 units of metal of the 1<sup>st</sup> type.\n- 2 units of metal of the 2<sup>nd</sup> type.\n- 2 units of metal of the 3<sup>rd</sup> type.\nIn total, we need 2 * 1 + 2 * 2 + 2 * 3 = 12 coins, which is smaller than or equal to budget = 15.\nNotice that we have 0 units of metal of each type and we have to buy all the required units of metal.\nIt can be proven that we can create at most 2 alloys.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> It is optimal to use the 2<sup>nd</sup> machine to create alloys.\nTo create 5 alloys we need to buy:\n- 5 units of metal of the 1<sup>st</sup> type.\n- 5 units of metal of the 2<sup>nd</sup> type.\n- 0 units of metal of the 3<sup>rd</sup> type.\nIn total, we need 5 * 1 + 5 * 2 + 0 * 3 = 15 coins, which is smaller than or equal to budget = 15.\nIt can be proven that we can create at most 5 alloys.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> It is optimal to use the 3<sup>rd</sup> machine to create alloys.\nTo create 2 alloys we need to buy the:\n- 1 unit of metal of the 1<sup>st</sup> type.\n- 1 unit of metal of the 2<sup>nd</sup> type.\nIn total, we need 1 * 5 + 1 * 5 = 10 coins, which is smaller than or equal to budget = 10.\nIt can be proven that we can create at most 2 alloys.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n, k <= 100</code></li>\n\t<li><code>0 <= budget <= 10<sup>8</sup></code></li>\n\t<li><code>composition.length == k</code></li>\n\t<li><code>composition[i].length == n</code></li>\n\t<li><code>1 <= composition[i][j] <= 100</code></li>\n\t<li><code>stock.length == cost.length == n</code></li>\n\t<li><code>0 <= stock[i] <= 10<sup>8</sup></code></li>\n\t<li><code>1 <= cost[i] <= 100</code></li>\n</ul>\n",
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"translatedTitle": "最大合金数",
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"translatedContent": "<p>假设你是一家合金制造公司的老板,你的公司使用多种金属来制造合金。现在共有 <code>n</code> 种不同类型的金属可以使用,并且你可以使用 <code>k</code> 台机器来制造合金。每台机器都需要特定数量的每种金属来创建合金。</p>\n\n<p>对于第 <code>i</code> 台机器而言,创建合金需要 <code>composition[i][j]</code> 份 <code>j</code> 类型金属。最初,你拥有 <code>stock[i]</code> 份 <code>i</code> 类型金属,而每购入一份 <code>i</code> 类型金属需要花费 <code>cost[i]</code> 的金钱。</p>\n\n<p>给你整数 <code>n</code>、<code>k</code>、<code>budget</code>,下标从 <strong>1</strong> 开始的二维数组 <code>composition</code>,两个下标从 <strong>1</strong> 开始的数组 <code>stock</code> 和 <code>cost</code>,请你在预算不超过 <code>budget</code> 金钱的前提下,<strong>最大化</strong> 公司制造合金的数量。</p>\n\n<p><strong>所有合金都需要由同一台机器制造。</strong></p>\n\n<p>返回公司可以制造的最大合金数。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]\n<strong>输出:</strong>2\n<strong>解释:</strong>最优的方法是使用第 1 台机器来制造合金。\n要想制造 2 份合金,我们需要购买:\n- 2 份第 1 类金属。\n- 2 份第 2 类金属。\n- 2 份第 3 类金属。\n总共需要 2 * 1 + 2 * 2 + 2 * 3 = 12 的金钱,小于等于预算 15 。\n注意,我们最开始时候没有任何一类金属,所以必须买齐所有需要的金属。\n可以证明在示例条件下最多可以制造 2 份合金。\n</pre>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]\n<strong>输出:</strong>5\n<strong>解释:</strong>最优的方法是使用第 2 台机器来制造合金。 \n要想制造 5 份合金,我们需要购买: \n- 5 份第 1 类金属。\n- 5 份第 2 类金属。 \n- 0 份第 3 类金属。 \n总共需要 5 * 1 + 5 * 2 + 0 * 3 = 15 的金钱,小于等于预算 15 。 \n可以证明在示例条件下最多可以制造 5 份合金。\n</pre>\n\n<p><strong class=\"example\">示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]\n<strong>输出:</strong>2\n<strong>解释:</strong>最优的方法是使用第 3 台机器来制造合金。\n要想制造 2 份合金,我们需要购买:\n- 1 份第 1 类金属。\n- 1 份第 2 类金属。\n总共需要 1 * 5 + 1 * 5 = 10 的金钱,小于等于预算 10 。\n可以证明在示例条件下最多可以制造 2 份合金。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= n, k <= 100</code></li>\n\t<li><code>0 <= budget <= 10<sup>8</sup></code></li>\n\t<li><code>composition.length == k</code></li>\n\t<li><code>composition[i].length == n</code></li>\n\t<li><code>1 <= composition[i][j] <= 100</code></li>\n\t<li><code>stock.length == cost.length == n</code></li>\n\t<li><code>0 <= stock[i] <= 10<sup>8</sup></code></li>\n\t<li><code>1 <= cost[i] <= 100</code></li>\n</ul>\n",
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