PAT 1037 Magic Coupon [模拟]

最新推荐文章于 2021-01-24 13:27:43 发布

原创最新推荐文章于 2021-01-24 13:27:43 发布 · 177 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

100 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，使用优惠券购买商品可以获得N倍的商品价值回馈。但是，如果优惠券N为正数并用于免费赠品，你需要支付N倍的赠品价值。任务是在每个优惠券和商品只能使用一次的情况下，制定策略以获得最大金额的回馈。

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:

-------------------------------------这是题目和解题的分割线-------------------------------------

先排序，再分正负区相加。

#include<cstdio>
#include<algorithm>

using namespace std;

//正数从大到小，负数从小到大 
bool cmp(int a,int b)
{
	if(a<0&&b<0) return a<b;
	return a>b;
}

int main()
{
	int n,m,i,j,a[100005],b[100005];
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	scanf("%d",&m);
	for(i=0;i<m;i++)
		scanf("%d",&b[i]);
	sort(a,a+n,cmp);
	sort(b,b+m,cmp);
	int backM = 0;
	//找到正负的边界下标 
	for(i=0;i<n;i++)
		if(a[i]<0) break;
	for(j=0;j<m;j++)
		if(b[j]<0) break;
	int x = min(i,j),x1 = i,x2 = j;
	//正数区 
	for(i=0;i<x;i++)
		backM += a[i]*b[i];
	//负数区 
	for(i=x1,j=x2;i<n,j<m;i++,j++)
	{
		if(a[i]*b[j]>0) backM += a[i]*b[j];
		else break;
	}
	printf("%d",backM);
	return 0;
}