Magic Coupon - 爱码网

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product… but hey, magically, they have some coupons with negative N’s!
For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.
Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.
输入描述：

Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 105, and it is guaranteed that all the numbers will not exceed 230.

输出描述：

For each test case, simply print in a line the maximum amount of money you can get back.

示例：

输入：
4
1 2 4 -1
4
7 6 -2 -3
输出：
43

题目分析：
在这道题中，它其实就是给你优惠券和产品，使用优惠券乘上你产品价值得到数，就是要返回你的金额；在题目中，开始的背景给你讲了优惠券为负，产品为正，得到的结果就是负，那么你就要给商店倒贴钱；反之，优惠券为正，产品价值为负，得到的结果也是负，也需要倒贴钱；但是在计算商店要返回给你的最大金额时，肯定不需要给店铺返钱呀！这是生活中的常识，所以我们就懂得了题目想要你回答什么。
你输入了一组优惠券和一组产品价值，要计算其得到金额的最大值，我们用输入示例来画图讲解：
Magic Coupon
代码如下：

import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        // 优惠券数量
        int numNC = sc.nextInt();
        // 优惠券价值
        long[] NC = new long[numNC];
        for(int i = 0; i < numNC; i++){
            NC[i] = sc.nextLong();
        }
        // 产品数量
        int numNP = sc.nextInt();
        // 产品价值
        long[] NP = new long[numNP];
        for(int i = 0; i < numNP; i++){
            NP[i] = sc.nextLong();
        }
        // 存放使用优惠券获得的金额
        long backMoney = 0;
        // 先排序
        Arrays.sort(NC);
        Arrays.sort(NP);
        // 遍历二维数组
        for(int i = 0, j = 0; i < numNC && j < numNP && NC[i] < 0 && NP[j] < 0; i++, j++){
            backMoney += NC[i] * NP[j];
        }
        for(int i = numNC-1, j = numNP-1; i > 0 && j > 0 && NC[i] > 0 && NP[j] > 0; i--, j--){
            backMoney += NC[i] * NP[j];
        }
        System.out.println(backMoney);
    }
}

本地IDEA中运行：
Magic Coupon
牛客网在线OJ运行：

注意： 在线OJ上的测试数据太大，记得要用long类型。