1037 Magic Coupon (25 分)

最新推荐文章于 2021-01-24 13:27:43 发布

莳绱枫少

最新推荐文章于 2021-01-24 13:27:43 发布

阅读量88

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT

本文链接：https://blog.youkuaiyun.com/qq_37866617/article/details/97428244

PAT 专栏收录该内容

76 篇文章

订阅专栏

在火星上的一家魔法商店，提供带有整数N的魔法优惠券和免费赠品，使用正N优惠券购买产品可获得N倍价值，但若用于赠品则需支付N倍价值。巧妙利用负N优惠券，目标是在遵守每个优惠券和产品仅能使用一次的规则下，最大化退款金额。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Input Specification:

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.

Output Specification:

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

Sample Input:

Sample Output:

#include <iostream>
#include <algorithm>
using namespace std;

int Coupon[100010], product[100010];

int main() {
	int n;
	
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &Coupon[i]);
	}
	
	int m;
	scanf("%d", &m);
	for (int i = 0; i < m; i++) {
		scanf("%d", &product[i]);
	}
	
	sort(Coupon, Coupon + n);
	sort(product, product + m);
	
	int i = 0;
	int long long ans = 0;
	while (i < n && i < m && Coupon[i] < 0 && product[i] < 0) {
		ans += (Coupon[i] * product[i]);
		i++;
	}
	
	i = n - 1;
	int j = m - 1;
	while (i >= 0 && j >= 0 && Coupon[i] > 0 && product[j] > 0) {
		ans += (Coupon[i] * product[j]);
		i--;
		j--;
	}
	
	printf("%d\n", ans);

	return 0;
}