1037 Magic Coupon (25 分)

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

 题目大意：给出两个数字序列，从这两个序列中分别选取相同数量的元素进行一对一相乘，问能得到的乘积之和最大为多少～
分析：把这两个序列都从小到大排序，将前面都是负数的数相乘求和，然后将后面都是正数的数相乘求和～

注意  有一组乘完是负数的  舍弃掉  不要了   这样的话  感觉排完序后 用在线处理  也可以   得到负值舍弃 直接往后走

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
	//freopen("in.txt", "r", stdin);
	int n, m,p,q,ans=0;

	cin >> n;
	vector<int> a(n); //初始化vector的时候最好 给vector的大小
	for (int i = 0; i < n; i++)
	{		
		cin >> a[i];
	}
	cin >> m;
	vector<int> b(m);
	for (int i = 0; i < m; i++)
	{
		cin >> b[i];
	}
	sort(a.begin(), a.end());
	sort(b.begin(), b.end());
	p = 0;
	q = 0;
	while (p<n&&q<m&&a[p]<0&&b[q]<0	)
	{
		ans += a[p] * b[q];
		p++;
		q++;
	}
	p = n - 1;
	q = m - 1;
	while (p>=0&&q>=0&&a[p]>0&&b[q]>0)
	{
		ans += a[p] * b[q];
		p--;
		q--;
	}
	cout << ans;
	//fclose(stdin);
	return 0;
}