[LeetCode] 1357. Apply Discount Every n Orders 每隔n个顾客打折
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 withn, thediscount, and theproductsand theirprices.double getBill(int[] product, int[] amount)Returns the final total of the bill with the discount applied (if any). Answers within10-5of the actual value will be accepted.
Example 1:
Input
["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]
[[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]]]
Output
[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]
Explanation
Cashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);
cashier.getBill([1,2],[1,2]); // return 500.0. 1st customer, no discount.
// bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]); // return 4000.0. 2nd customer, no discount.
// bill = 10 * 300 + 10 * 100 = 4000.
cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]); // return 800.0. 3rd customer, 50% discount.
// Original bill = 1600
// Actual bill = 1600 * ((100 - 50) / 100) = 800.
cashier.getBill([4],[10]); // return 4000.0. 4th customer, no discount.
cashier.getBill([7,3],[10,10]); // return 4000.0. 5th customer, no discount.
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0. 6th customer, 50% discount.
// Original bill = 14700, but with
// Actual bill = 14700 * ((100 - 50) / 100) = 7350.
cashier.getBill([2,3,5],[5,3,2]); // return 2500.0. 7th customer, no discount.
Constraints:
1 <= n <= 1040 <= discount <= 1001 <= products.length <= 200prices.length == products.length1 <= products[i] <= 2001 <= prices[i] <= 1000- The elements in
productsare unique. 1 <= product.length <= products.lengthamount.length == product.lengthproduct[j]exists inproducts.1 <= amount[j] <= 1000- The elements of
productare unique. - At most
1000calls will be made togetBill. - Answers within
10-5of the actual value will be accepted.
这道题让给每n个订单打折，给了个折扣数，以及产品和价格数组，表示 products[i] 产品的价格为 prices[i]，现在给了一个产品组 product 和数量数组 amount，表示购买 product[i] 产品的数量为 ampunt[i] 个，让求给定的订单的价格。这道题没有太大的难度，先用一个 HashMap 来简历产品和其价格之间的映射，然后用个全局变量 cnt 来统计已接受订单的个数。在 getBill 函数中，先计算出买所有订单内的产品的总价，然后判断当前的订单个数 cnt 是否能整除n，能的话再计算出打折后的价格即可，参见代码如下：
class Cashier {
public:
Cashier(int n, int discount, vector<int>& products, vector<int>& prices) {
this->n = n;
this->discount = discount;
this->cnt = 0;
for (int i = 0; i < products.size(); ++i) {
m[products[i]] = prices[i];
}
}
double getBill(vector<int> product, vector<int> amount) {
double res = 0;
for (int i = 0; i < product.size(); ++i) {
res += m[product[i]] * amount[i];
}
if (++cnt % n == 0) {
res = (res * (100 - discount)) / 100;
}
return res;
}
private:
int n, discount, cnt;
unordered_map<int, int> m;
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1357
类似题目：
Apply Discount to Prices
参考资料：
https://leetcode.com/problems/apply-discount-every-n-orders
