PAT A1037 Magic Coupon (25分)

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805451374313472

题目描述
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.

输入
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​​ .

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

样例输入
4
1 2 4 -1
4
7 6 -2 -3

样例输出
43

代码

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

const int maxn = 100010;
int coupon[maxn], product[maxn];
int main(){
	int n, m;
	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);
	sort(product, product + m);
	int i = 0, j, ans = 0;
	while(i < n && i < m && coupon[i] < 0 && product[i] < 0) {
		ans += coupon[i] * product[i];
		i++;
	}
	i = n - 1;
	j = m - 1;
	while(i >= 0 && j >= 0 && coupon[i] > 0 && product[j] > 0) {
		ans += coupon[i] * product[j];
		i--;
		j--;
	}
	printf("%d\n", ans);
	return 0;
}