PAT甲级 1037. Magic Coupon (25)

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本文介绍了一种算法，用于解决如何使用特定数值的优惠券与商品价值匹配以获得最大收益的问题。通过将正数与正数、负数与负数进行配对，并确保优惠券与商品各不重复使用。

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:

Sample Output:

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;
}