题目地址
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

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=100010,inf=100000000;
vector<int>pos1,pos2,neg1,neg2;
bool cmp(int a,int b){
    return a>b;
}
int main(){
    int m,n,num;
    cin>>m;
    for(int i=0;i<m;i++){
        cin>>num;
        if(num>=0) pos1.push_back(num);
        else neg1.push_back(num);
    }
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>num;
        if(num>=0) pos2.push_back(num);
        else neg2.push_back(num);
    }
    sort(pos1.begin(),pos1.end(),cmp);sort(pos2.begin(),pos2.end(),cmp);
    sort(neg1.begin(),neg1.end());sort(neg2.begin(),neg2.end());
    int i=0,j=0,ans=0;
    while(i<neg1.size()&&i<neg2.size()){
        ans+=neg1[i]*neg2[i];i++;
    }
    while(j<pos1.size()&&j<pos2.size()){
        ans+=pos1[j]*pos2[j];j++;
    }
    cout<<ans;
}