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 − }, and a set of product values { 7 6 − − } (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, 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
题目分析：排序就好

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 bool compare(const int& a, const int& b)
 {
     return a < b;
 }
 int main()
 {
     int NC, NP;
     int total = ;
     cin >> NC;
     vector<int> C(NC);
     for (int i = ; i < NC; i++)
         cin >> C[i];
     cin >> NP;
     vector<int>P(NP);
     for (int i = ; i < NP; i++)
         cin >> P[i];
     sort(C.begin(), C.end(), compare);
     sort(P.begin(), P.end(), compare);
     auto it1 = C.begin();
     auto it2 = P.begin();
     for(; it1!=C.end()&&it2!=P.end()&&*it1<&&*it2<;)
     {
         total += (*it1) * (*it2);
         it1++;
         it2++;
     }
     it1 = (C.end() - );
     it2 = (P.end() - );
     while (it1>=C.begin() &&it2>= P.begin()&&*it1>&&*it2>)
     {
         total += (*it1) * (*it2);
         it1--;
         it2--;
     }
     cout << total;
     return ;
 }