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.

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​​ ,NP​​ ≤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

思路

将优惠券和商品进行排序，从都是负数的开始相乘相加，都是正数的末端开始相乘相加。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int c, p;
    scanf("%d", &c);
    vector<int> coup(c);
    for (int i = 0; i < c; i++) scanf("%d", &coup[i]);
    scanf("%d", &p);
    vector<int> pro(p);
    for (int i = 0; i < p; i++) scanf("%d", &pro[i]);
    sort(coup.begin(), coup.end());
    sort(pro.begin(), pro.end());
    int sum = 0, cl = 0, cr = coup.size() - 1, pl = 0, pr = pro.size() - 1;
    while (cl < c && pl < p && coup[cl] < 0 && pro[pl] < 0) {
        sum += coup[cl] * pro[pl];
        cl++; pl++;
    }
    while (cr >= 0 && pr >= 0 && coup[cr] > 0 && pro[pr] > 0) {
        sum += coup[cr] * pro[pr];
        cr--; pr--;
    }
    printf("%d", sum);
    return 0;
}