You are the owner of a company that creates alloys using various types of metals. There are n different types of metals available, and you have access to k machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy.
For the ith machine to create an alloy, it needs composition[i][j] units of metal of type j. Initially, you have stock[i] units of metal type i, and purchasing one unit of metal type i costs cost[i] coins.
Given integers n, k, budget, a 1-indexed 2D array composition, and 1-indexed arrays stock and cost, your goal is to maximize the number of alloys the company can create while staying within the budget of budget coins.
All alloys must be created with the same machine.
\n\nReturn the maximum number of alloys that the company can create.
\n\n\n
Example 1:
\n\n\nInput: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]\nOutput: 2\nExplanation: It is optimal to use the 1st machine to create alloys.\nTo create 2 alloys we need to buy the:\n- 2 units of metal of the 1st type.\n- 2 units of metal of the 2nd type.\n- 2 units of metal of the 3rd 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\n\n
Example 2:
\n\n\nInput: n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]\nOutput: 5\nExplanation: It is optimal to use the 2nd machine to create alloys.\nTo create 5 alloys we need to buy:\n- 5 units of metal of the 1st type.\n- 5 units of metal of the 2nd type.\n- 0 units of metal of the 3rd 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\n\n
Example 3:
\n\n\nInput: n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]\nOutput: 2\nExplanation: It is optimal to use the 3rd machine to create alloys.\nTo create 2 alloys we need to buy the:\n- 1 unit of metal of the 1st type.\n- 1 unit of metal of the 2nd 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\n\n
\n
Constraints:
\n\n1 <= n, k <= 1000 <= budget <= 108composition.length == kcomposition[i].length == n1 <= composition[i][j] <= 100stock.length == cost.length == n0 <= stock[i] <= 1081 <= cost[i] <= 100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 n 种不同类型的金属可以使用，并且你可以使用 k 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。
对于第 i 台机器而言，创建合金需要 composition[i][j] 份 j 类型金属。最初，你拥有 stock[i] 份 i 类型金属，而每购入一份 i 类型金属需要花费 cost[i] 的金钱。
给你整数 n、k、budget，下标从 1 开始的二维数组 composition，两个下标从 1 开始的数组 stock 和 cost，请你在预算不超过 budget 金钱的前提下，最大化 公司制造合金的数量。
所有合金都需要由同一台机器制造。
\n\n返回公司可以制造的最大合金数。
\n\n\n\n
示例 1：
\n\n\n输入：n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]\n输出：2\n解释：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\n\n
示例 2：
\n\n\n输入：n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]\n输出：5\n解释：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\n\n
示例 3：
\n\n\n输入：n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]\n输出：2\n解释：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\n\n
\n\n
提示：
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