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 NCN_C, followed by a line with NCN_C coupon integers. Then the next line contains the number of products , NPN_Pfollowed by a line with NPN_P product values. Here 1≤NC,NP≤1051≤N_C,N_P≤10^5, and it is guaranteed that all the numbers will not exceed 2302^{30}.

Output Specification:

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

Sample Input:

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1 2 4 -1
4
7 6 -2 -3

Sample Output:

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43

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。
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num_a = int(input())
a = list(map(int, input().split()))
a_pos = sorted([i for i in a if i > 0], reverse=True)
a_neg = sorted([i for i in a if i < 0])
num_b = int(input())
b = list(map(int, input().split()))
b_pos = sorted([i for i in b if i > 0], reverse=True)
b_neg = sorted([i for i in b if i < 0])
print(sum([i * j for i, j in zip(a_pos, b_pos)]) + sum([i * j for i, j in zip(a_neg, b_neg)]))