1037 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

题意：

　　给出两个数组，求出这两个数组中，任意选取一个数字的乘积和的最大值。（每个数字只能使用一次）

思路：

　　简单模拟。

Code：

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 int main() {
 6     int nc, np;
 7     cin >> nc;
 8     vector<int> coupons(nc);
 9     for (int i = 0; i < nc; ++i) cin >> coupons[i];
10     sort(coupons.begin(), coupons.end(), greater<int>());
11     cin >> np;
12     vector<int> products(np);
13     for (int i = 0; i < np; ++i) cin >> products[i];
14     sort(products.begin(), products.end(), greater<int>());
15     vector<bool> couponsUsed(nc + 1, false);
16     vector<bool> productsUsed(np + 1, false);
17     int i = 0, j = 0, res = 0;
18     while (i < nc && j < np) {
19         if (coupons[i] * products[j] > 0) {
20             couponsUsed[i] = true;
21             productsUsed[j] = true;
22             res += coupons[i++] * products[j++];
23         } else
24             break;
25     }
26     if (i == nc || j == np)
27         cout << res << endl;
28     else {
29         i = nc - 1;
30         j = np - 1;
31         while (i >= 0 && j >= 0) {
32             if (couponsUsed[i] || productsUsed[j]) break;
33             if (coupons[i] * products[j] > 0) {
34                 res += coupons[i--] * products[j--];
35             } else
36                 break;
37         }
38         cout << res << endl;
39     }
40     return 0;
41 }