PAT甲级 1037. Magic Coupon (25)

题目：

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:

41 2 4 -147 6 -2 -3

Sample Output:

43

思路：

这道题应该就是正数与正数相乘，负数与负数相乘。这里，我写的时候把vector里的sort记成降序排序了，但这是升序排序，这个要注意。

代码：

#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){int NC, NP;cin >> NC;vector<int>coupon(NC);int i;for (i = 0; i < NC; ++i){cin >> coupon[i];}cin >> NP;vector<int>values(NP);for (i = 0; i < NP; ++i){cin >> values[i];}sort(coupon.begin(),coupon.end());sort(values.begin(),values.end());int s1 = 0;int s2 = 0;while (s1<NC && coupon[s1]<0){++s1;}while (s2<NP && values[s2]<0){++s2;}int money = 0;int j;i = j = 0;while (i < s1 && j < s2){money += coupon[i] * values[j];++i; ++j;}i = NC - 1;j = NP - 1;while (i >= s1 && j >= s2){money += coupon[i] * values[j];--i; --j;}cout << money << endl;system("pause");return 0;}