1037 Magic Coupon (25point(s))

问题

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 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≤N​C​​,N​P​​≤10​5​​, 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

解决方法

分析：贪心算法

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
	long long n1, n2,p=0,q=0,result=0;
	scanf("%lld", &n1);
	vector<int>v1(n1);
	for (int i = 0; i < n1; i++) scanf("%lld", &v1[i]);
	scanf("%lld", &n2);
	vector<int>v2(n2);
	for (int i = 0; i < n2; i++) scanf("%lld", &v2[i]);
	sort(v1.begin(), v1.end());
	sort(v2.begin(), v2.end());
	while (p < n1&&q < n2&&v1[p] < 0 && v2[q] < 0)
	{
		result += v1[p] * v2[q];
		p++;
		q++;
	}
	p = n1 - 1;
	q = n2 - 1;
	while (p < n1&&q < n2&&v1[p]>0 && v2[q]>0)
	{
		result += v1[p] * v2[q];
		p--;
		q--;
	}
	printf("%lld", result);
	return 0;
}

后记

其中的元素不一定用完