PAT 1037. Magic Coupon (25)

1037. Magic Coupon (25)

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 -1}, and a set of product values {7 6 -2 -3} (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<= NC, NP <= 105, 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

应用排序不等式。将数字分为整数和负数，将最大（正数）的优惠券匹配价值最大（正数）的产品，最小（负数）匹配最小（负数）价值的产品。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 bool cmp(int integer1, int integer2)
 9 {
10     return integer1 > integer2;
11 }
12 
13 int main()
14 {
15     int NC, NP;
16     vector<int> positiveCoupons, negativeCoupons,  positiveProduct, negativeProduct;
17     cin >> NC;
18     for (int i = 0; i < NC; i++)
19     {
20         int coupon;
21         scanf("%d", &coupon);
22         if (coupon > 0)
23             positiveCoupons.push_back(coupon);
24         else if (coupon < 0)
25             negativeCoupons.push_back(coupon);
26 
27     }
28     sort(positiveCoupons.begin(), positiveCoupons.end(), cmp);
29     sort(negativeCoupons.begin(), negativeCoupons.end());
30     cin >> NP;
31     for (int i = 0; i < NP; i++)
32     {
33         int product;
34         scanf("%d", &product);
35         if (product > 0)
36             positiveProduct.push_back(product);
37         else if (product < 0)
38             negativeProduct.push_back(product);
39 
40     }
41     sort(positiveProduct.begin(), positiveProduct.end(), cmp);
42     sort(negativeProduct.begin(), negativeProduct.end());
43 
44     int MaxMonmey = 0;
45     for (int i = 0; i < positiveCoupons.size() && i < positiveProduct.size(); i++)
46         MaxMonmey += positiveCoupons[i] * positiveProduct[i];
47     for (int i = 0; i < negativeCoupons.size() && negativeProduct.size(); i++)
48         MaxMonmey += negativeCoupons[i] * negativeProduct[i];
49 
50     cout << MaxMonmey;
51 }

 

posted @ 2015-08-11 14:03  JackWang822  阅读(152)  评论(0)  编辑  收藏  举报