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 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
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 }