PAT甲级——A1037 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 − }, 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

 #include <iostream>
 #include <vector>
 #include <string>
 #include <algorithm>
 using namespace std;
 //牛客这道题涉及数相乘会int溢出，而PAT没有，所以pat使用int类型就可以，而牛客需要使用longlong型
 long long Nc, Np, p = , q = , a, res = ;
 int main()
 {
     cin >> Nc;
     vector<long>Vc, Vp;
     for (int i = ; i < Nc; ++i)
     {
         cin >> a;
         Vc.push_back(a);
     }
     cin >> Np;
     for (int i = ; i < Np; ++i)
     {
         cin >> a;
         Vp.push_back(a);
     }
     sort(Vc.begin(), Vc.end(), [](long long u, long long v) {return u < v; });
     sort(Vp.begin(), Vp.end(), [](long long u, long long v) {return u < v; });
     while (p < Nc&&q < Np&&Vc[p] <  && Vp[q] < )//先负数相乘
     {
         res += Vc[p] * Vp[q];
         ++p;
         ++q;
     }
     p = Nc - ;
     q = Np - ;
     while (p >=  & q >=  && Vc[p] >  && Vp[q] > )//正数从后开始乘
     {
         res += Vc[p] * Vp[q];
         --p;
         --q;
     }
     cout << res << endl;
     return ;
 }