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 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, 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
1 #include <iostream> 2 #include <vector> 3 #include <string> 4 #include <algorithm> 5 using namespace std; 6 //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 7 long long Nc, Np, p = 0, q = 0, a, res = 0; 8 int main() 9 { 10 cin >> Nc; 11 vector<long>Vc, Vp; 12 for (int i = 0; i < Nc; ++i) 13 { 14 cin >> a; 15 Vc.push_back(a); 16 } 17 cin >> Np; 18 for (int i = 0; i < Np; ++i) 19 { 20 cin >> a; 21 Vp.push_back(a); 22 } 23 sort(Vc.begin(), Vc.end(), [](long long u, long long v) {return u < v; }); 24 sort(Vp.begin(), Vp.end(), [](long long u, long long v) {return u < v; }); 25 while (p < Nc&&q < Np&&Vc[p] < 0 && Vp[q] < 0)//先负数相乘 26 { 27 res += Vc[p] * Vp[q]; 28 ++p; 29 ++q; 30 } 31 p = Nc - 1; 32 q = Np - 1; 33 while (p >= 0 & q >= 0 && Vc[p] > 0 && Vp[q] > 0)//正数从后开始乘 34 { 35 res += Vc[p] * Vp[q]; 36 --p; 37 --q; 38 } 39 cout << res << endl; 40 return 0; 41 }