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

第一行：nc（优惠券的个数）
第二行：nc个优惠券的整数
第三行：np（产品的个数）
第四行：np个产品价值的整数

输出：

能得到的钱的最大值

解题思路

本题实际上当优惠券值c和产品价值p
的符号一致时能得到pc的钱，如果符号不一致则需要向商家支付|pc|。因此将优惠券和产品价值按正负值分别存入数组,然后按绝对值排序，将产品的正（负）值数组和优惠券的正（负）值数组值挨个相乘得到值v，累加v，得到结果。

代码

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
static int cmp1(int a, int b) {
    return a >b;
}
int main() {
    int nc, np;
    vector<int>pcoupons, ncoupons,pproduct,nproduct;
    int t;
    scanf("%d", &nc);
    for (int i = 0; i < nc; i++) {
        scanf("%d", &t);
        if (t > 0)pcoupons.push_back(t);
        else if(t<0){
            ncoupons.push_back(t);
        }
    }
    scanf("%d", &np);
    for (int i = 0; i < np; i++) {
        scanf("%d", &t);
        if (t > 0)pproduct.push_back(t);
        else if (t<0) {
            nproduct.push_back(t);
        }
    }
    sort(pcoupons.begin(), pcoupons.end(), cmp1);
    sort(pproduct.begin(), pproduct.end(), cmp1);
    sort(nproduct.begin(), nproduct.end());
    sort(ncoupons.begin(), ncoupons.end());
    int res = 0;
    for (int i = 0; i < pcoupons.size() && i < pproduct.size(); i++) {
        res += pproduct[i] * pcoupons[i];
    }
    for (int i = 0; i < ncoupons.size() && i < nproduct.size(); i++) {
        res += ncoupons[i] * nproduct[i];
    }
    printf("%d", res);
    return 0;
}