1357. Apply Discount Every n Orders

There is a sale in a supermarket, there will be a discount every n customer.
There are some products in the supermarket where the id of the i-th product is products[i] and the price per unit of this product is prices[i].
The system will count the number of customers and when the n-th customer arrive he/she will have a discount on the bill. (i.e if the cost is x the new cost is x - (discount * x) / 100). Then the system will start counting customers again.
The customer orders a certain amount of each product where product[i] is the id of the i-th product the customer ordered and amount[i] is the number of units the customer ordered of that product.

Implement the Cashier class:

Cashier(int n, int discount, int[] products, int[] prices) Initializes the object with n, the discount, the products and their prices.
double getBill(int[] product, int[] amount) returns the value of the bill and apply the discount if needed. Answers within 10^-5 of the actual value will be accepted as correct.

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, bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]);                      // return 4000.0
cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]);    // return 800.0, The bill was 1600.0 but as this is the third customer, he has a discount of 50% which means his bill is only 1600 - 1600 * (50 / 100) = 800.
cashier.getBill([4],[10]);                           // return 4000.0
cashier.getBill([7,3],[10,10]);                      // return 4000.0
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0, Bill was 14700.0 but as the system counted three more customers, he will have a 50% discount and the bill becomes 7350.0
cashier.getBill([2,3,5],[5,3,2]);                    // return 2500.0

Constraints:

1 <= n <= 10^4
0 <= discount <= 100
1 <= products.length <= 200
1 <= products[i] <= 200
There are not repeated elements in the array products.
prices.length == products.length
1 <= prices[i] <= 1000
1 <= product.length <= products.length
product[i] exists in products.
amount.length == product.length
1 <= amount[i] <= 1000
At most 1000 calls will be made to getBill.
Answers within 10^-5 of the actual value will be accepted as correct.

题目出这么长的leetcode实属人间之屑

题意是设计一个class，能初始化cashier和getbill，每n个人有一个可以获得discount

class Cashier {
    public int mod = 0;
    int n = 0;
    double disc = 0;
    private int[] products;
    private int[] prices;
    public Cashier(int n, int discount, int[] products, int[] prices) {
        this.n = n;
        disc = discount;
        this.products = products;
        this.prices = prices;
    }
    
    public double getBill(int[] product, int[] amount) {
        mod ++;
        double bill = 0;
        for(int i = 0; i < product.length; i++){
            for(int j = 0; j < products.length; j++){
                if(products[j] == product[i]) bill += prices[j] * amount[i];
            }
            
        }
        return mod % n == 0 ? (bill*(1 - disc / 100)) : bill;
    }
}

方法一：一时大意忘了hashmap这个东西，就感觉怪怪的

class Cashier {
    private int mod = 0;
    private Map<Integer, Integer> map;
    private double disc = 0;
    private int n = 0;
    public Cashier(int n, int discount, int[] products, int[] prices) {
        this.n = n;
        disc = discount;
        this.map = new HashMap();
        for(int i = 0; i < products.length; i++) map.put(products[i], prices[i]);
    }
    
    public double getBill(int[] product, int[] amount) {
        mod ++;
        double bill = 0;
        for(int i = 0; i < product.length; i++){
            bill += map.get(product[i]) * amount[i];
        }
        return mod % n == 0 ? (bill*(1 - disc / 100)) : bill;
    }
}

用hashmap来记录products和prices