1.题目给出一些coupon和product，求最大的value。
2.最大的正数coupon与最大的正数product相乘。
3.最小的负数coupon与最小的负数product相乘（最小的负数即绝对值最大）。
4.其他没有匹配的就不购买了。
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
AC代码：
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
bool cmp(const int&a, const int&b)
{
return a > b;
}
int main(void)
{
int n;
vector<int> couponP(0);
vector<int> couponN(0);
vector<int> productP(0);
vector<int> productN(0);
cin >> n;
for (int i = 0; i < n; i++)
{
int a;
scanf("%d", &a);
if (a < 0) couponN.push_back(a);
else couponP.push_back(a);
}
cin >> n;
for (int i = 0; i < n; i++)
{
int a;
scanf("%d", &a);
if (a < 0) productN.push_back(a);
else productP.push_back(a);
}
sort(couponP.begin(), couponP.end(), cmp);
sort(couponN.begin(), couponN.end());
sort(productP.begin(), productP.end(), cmp);
sort(productN.begin(), productN.end());
int sum = 0;
for (int i = 0; i < couponP.size() && i < productP.size(); i++)
sum += couponP[i] * productP[i];
for (int i = 0; i < couponN.size() && i < productN.size(); i++)
sum += couponN[i] * productN[i];
cout << sum << endl;
return 0;
}