A1037 Magic Coupon

1037 Magic Coupon （25 分)

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​​≤​​, and it is guaranteed that all the numbers will not exceed ​​.

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

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

int main()
{
	int nc, np;
	scanf("%d", &nc); 
	long long coupon[nc];
	for(int i=0; i<nc; i++){
		scanf("%lld", &coupon[i]);
	} 
	scanf("%d", &np);
	long long product[np];
	for(int i=0; i<np; i++){
		scanf("%lld", &product[i]);
	}
	sort(coupon, coupon + nc);
	sort(product, product + np);
	long long ans = 0;
	for(int i=0; i<nc && i<np && coupon[i] < 0 && product[i] < 0; i++){
		ans += coupon[i] * product[i];
	}
	for(int i=nc-1, j=np-1; i >= 0 && j >= 0 && coupon[i] > 0 && product[j] > 0; i--, j--){
		ans +=coupon[i] * product[j];
	}
	printf("%lld", ans);
	return 0;
}