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 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:

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

Sample Output:

43

在输入的时候就把两组数据中的正负数分开存放，正数从大到小排序，负数从小到大排序，正数和正数相乘，负数和负数相乘，再加起来即可。 

#include <iostream>
#include <algorithm>
using namespace std;
long long n1, n2, ans, ax, bx, ay, by; //x代表正数个数，y代表负数个数
long long a[100010], b[100010], a1[100010], b1[100010];

int main() {
	cin >> n1;
	long long t;
	for (int i = 1; i <= n1; i++) {
		cin >> t;
		if (t > 0) {
			a[++ax] = t;
		}
		if (t < 0) {
			a1[++ay] = t;
		}
	}
	cin >> n2;
	for (int i = 1; i <= n2; i++) {
		cin >> t;
		if (t > 0) {
			b[++bx] = t;
		}
		if (t < 0) {
			b1[++by] = t;
		}
	}
	sort(a + 1, a + 1 + ax, greater<int>());
	sort(b + 1, b + 1 + bx, greater<int>());
	sort(a1 + 1, a1 + 1 + ay);
	sort(b1 + 1, b1 + 1 + by);
	int i = 1, j = 1;
	while (i <= ax && j <= bx) {
		ans += a[i] * b[j];
		i++;
		j++;
	}
	i = 1, j = 1;
	while (i <= ay && j <= by) {
		ans += a1[i] * b1[j];
		i++;
		j++;
	}
	cout << ans;
	return 0;
}