# 题目

he 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.

### Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons $N_C$, followed by a line with $N_C$ coupon integers. Then the next line contains the number of products , $N_P$followed by a line with $N_P$ product values. Here $1≤N_C,N_P≤10^5$, and it is guaranteed that all the numbers will not exceed $2^{30}$.

### Output Specification:

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

# 题解

## 思路

• 题目说了一大堆，其实题目无非就是

• 给你两个序列
• 从第一个序列挑数乘以第二序列挑一个数
• 使这些乘积最大
• 数选了一次就不能选，但是可以一次都不选
• 因为正的越大乘以正的越大，效果越好，负的越大乘以负的越大效果越好，而一正一负就不乘了。

• 那我们把两个序列按照正负分为四个序列，一正一负二正二负，同时排序

• 利用zip和sum，对两个正序列的值相乘，对两个负序列的值相乘，并加起来

## 数据结构

• a 是第一个序列
• a_pos 是第一个序列的正数
• a_neg 是第一个序列的负数
• b 是第一个序列
• b_pos 是第一个序列的正数
• b_neg 是第一个序列的负数

## 算法

• 对a_pos,b_pos按从大到小排序
• 对a_neg,b_neg按从小到大排序
• 将a_pos,b_pos用zip组合起来，对组合起来的每一项，求它们的乘积，并把所有乘积加起来。
• 对a_neg,b_neg同理

## 代码

• 因为使用Python就能AC，因此只使用了Python的代码。