PAT A1037. Magic Coupon (25)

最新推荐文章于 2021-10-05 15:56:52 发布

jolivan

最新推荐文章于 2021-10-05 15:56:52 发布

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分类专栏： PAT 文章标签：贪心算法

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本文介绍了一种在火星魔法店利用特殊优惠券获取最大利益的算法。通过将正数优惠券与正数商品价值匹配，并利用负数优惠券与负数商品价值结合，实现了利益最大化。

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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 <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
#include <string>
#define Max 100000
using namespace std;
bool cmp(int a,int b)
{
	return a>b;
}
int main()
{
	int n,m;
	int S[Max],K[Max];
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&S[i]);
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d",&K[i]);
	}
	sort(S,S+n,cmp);
	sort(K,K+m,cmp);
	int num=0,f;
	for(int i=0;i<n&&i<m;i++)
	{
		if(S[i]>0&&K[i]>0)
		{
			num+=S[i]*K[i];
			
		}
		else 
		{
			f=i;
			break;
		}
	}
	for(int i=n-1;i>=f;i--)
	{
		for(int j=m-1;j>=f;j--)
		{
			m=m-1;
			if(S[i]<0&&K[j]<0)
			{
				num+=S[i]*K[j];
				
				break;
			}
		}
	}
	printf("%d\n",num);
    system("pause");
	return 0;
}