PAT (Advanced Level) Practice 1037 Magic Coupon

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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 N​C​​, followed by a line with N​C​​ coupon integers. Then the next line contains the number of products N​P​​, followed by a line with N​P​​ product values. Here 1≤N​C​​,N​P​​≤10​5​​, and it is guaranteed that all the numbers will not exceed 2​30​​.

Output Specification:

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

Sample Input:

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

Sample Output:

43

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;
}

 

result：