PAT (Advanced Level) Practice 1037 Magic Coupon

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在火星的一家魔法商店中，玩家需要通过合理使用不同数值的魔法优惠券和选择产品来最大化自己的收益。优惠券可以将产品的价值放大N倍，但若应用于免费赠品，则需支付N倍的价值。玩家的目标是在有限的优惠券和产品中，找到最佳组合以获取最大回报。

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1037 Magic Coupon

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:

running code：

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100010;
int coupon[maxn],product[maxn];
int main()
{
	int m,n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&coupon[i]);
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d",&product[i]);
	}
	sort(coupon,coupon+n);//从小到大排序
	sort(product,product+m);//从小到大排序
	int i=0,j,ans=0;
	while(i<n&&i<m&&coupon[i]<0&&product[i]<0)//负数集合相乘累加
	{
		ans+=coupon[i]*product[i];
		i++;
	}
	i=n-1;
	j=m-1;
	while(i>=0&&j>=0&&coupon[i]>0&&product[j]>0)//正数集合相乘累加
	{
		ans+=coupon[i]*product[j];
		i--;
		j--;
	}
	printf("%d\n",ans);
	return 0;
}