{ "data": { "question": { "questionId": "1459", "questionFrontendId": "1357", "categoryTitle": "Algorithms", "boundTopicId": 102331, "title": "Apply Discount Every n Orders", "titleSlug": "apply-discount-every-n-orders", "content": "
There is a supermarket that is frequented by many customers. The products sold at the supermarket are represented as two parallel integer arrays products
and prices
, where the ith
product has an ID of products[i]
and a price of prices[i]
.
When a customer is paying, their bill is represented as two parallel integer arrays product
and amount
, where the jth
product they purchased has an ID of product[j]
, and amount[j]
is how much of the product they bought. Their subtotal is calculated as the sum of each amount[j] * (price of the jth product)
.
The supermarket decided to have a sale. Every nth
customer paying for their groceries will be given a percentage discount. The discount amount is given by discount
, where they will be given discount
percent off their subtotal. More formally, if their subtotal is bill
, then they would actually pay bill * ((100 - discount) / 100)
.
Implement the Cashier
class:
Cashier(int n, int discount, int[] products, int[] prices)
Initializes the object with n
, the discount
, and the products
and their prices
.double getBill(int[] product, int[] amount)
Returns the final total of the bill with the discount applied (if any). Answers within 10-5
of the actual value will be accepted.\n
Example 1:
\n\n\nInput\n["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]\n[[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]]\nOutput\n[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]\nExplanation\nCashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);\ncashier.getBill([1,2],[1,2]); // return 500.0. 1st customer, no discount.\n // bill = 1 * 100 + 2 * 200 = 500.\ncashier.getBill([3,7],[10,10]); // return 4000.0. 2nd customer, no discount.\n // bill = 10 * 300 + 10 * 100 = 4000.\ncashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]); // return 800.0. 3rd customer, 50% discount.\n // Original bill = 1600\n // Actual bill = 1600 * ((100 - 50) / 100) = 800.\ncashier.getBill([4],[10]); // return 4000.0. 4th customer, no discount.\ncashier.getBill([7,3],[10,10]); // return 4000.0. 5th customer, no discount.\ncashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0. 6th customer, 50% discount.\n // Original bill = 14700, but with\n // Actual bill = 14700 * ((100 - 50) / 100) = 7350.\ncashier.getBill([2,3,5],[5,3,2]); // return 2500.0. 6th customer, no discount.\n\n\n
\n
Constraints:
\n\n1 <= n <= 104
0 <= discount <= 100
1 <= products.length <= 200
prices.length == products.length
1 <= products[i] <= 200
1 <= prices[i] <= 1000
products
are unique.1 <= product.length <= products.length
amount.length == product.length
product[j]
exists in products
.1 <= amount[j] <= 1000
product
are unique.1000
calls will be made to getBill
.10-5
of the actual value will be accepted.超市里正在举行打折活动,每隔 n
个顾客会得到 discount
的折扣。
超市里有一些商品,第 i
种商品为 products[i]
且每件单品的价格为 prices[i]
。
结账系统会统计顾客的数目,每隔 n
个顾客结账时,该顾客的账单都会打折,折扣为 discount
(也就是如果原本账单为 x
,那么实际金额会变成 x - (discount * x) / 100
),然后系统会重新开始计数。
顾客会购买一些商品, product[i]
是顾客购买的第 i
种商品, amount[i]
是对应的购买该种商品的数目。
请你实现 Cashier
类:
Cashier(int n, int discount, int[] products, int[] prices)
初始化实例对象,参数分别为打折频率 n
,折扣大小 discount
,超市里的商品列表 products
和它们的价格 prices
。double getBill(int[] product, int[] amount)
返回账单的实际金额(如果有打折,请返回打折后的结果)。返回结果与标准答案误差在 10^-5
以内都视为正确结果。\n\n
示例 1:
\n\n输入\n["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]\n[[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]]\n输出\n[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]\n解释\nCashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);\ncashier.getBill([1,2],[1,2]); // 返回 500.0, 账单金额为 = 1 * 100 + 2 * 200 = 500.\ncashier.getBill([3,7],[10,10]); // 返回 4000.0\ncashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]); // 返回 800.0 ,账单原本为 1600.0 ,但由于该顾客是第三位顾客,他将得到 50% 的折扣,所以实际金额为 1600 - 1600 * (50 / 100) = 800 。\ncashier.getBill([4],[10]); // 返回 4000.0\ncashier.getBill([7,3],[10,10]); // 返回 4000.0\ncashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // 返回 7350.0 ,账单原本为 14700.0 ,但由于系统计数再次达到三,该顾客将得到 50% 的折扣,实际金额为 7350.0 。\ncashier.getBill([2,3,5],[5,3,2]); // 返回 2500.0\n\n\n
\n\n
提示:
\n\n1 <= n <= 10^4
0 <= discount <= 100
1 <= products.length <= 200
1 <= products[i] <= 200
products
列表中 不会 有重复的元素。prices.length == products.length
1 <= prices[i] <= 1000
1 <= product.length <= products.length
product[i]
在 products
出现过。amount.length == product.length
1 <= amount[i] <= 1000
1000
次对 getBill
函数的调用。10^-5
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