1037 Magic Coupon 贪心

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 − }, and a set of product values { 7 6 − − } (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, 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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm> 
#define ll long long
#define inf  100000
using namespace std;
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    ll n,m,a[inf],b[inf],sum=0;
    cin>>n;
    for(int i=0;i<n;i++)
    cin>>a[i];
    cin>>m;
    for(int i=0;i<m;i++)
    cin>>b[i];
    sort(a,a+n,cmp);
    sort(b,b+m,cmp);
    int i=0,j=0;
    while(i<n||j<m)
    {
        if(a[i]>0&&b[j]>0)
        {
        sum=sum+a[i]*b[j];
        }
        i++;
        j++;
    }
    i=n-1;
    j=m-1;
    while(i>=0||j>=0)
    {
        if(a[i]<0&&b[j]<0)
        {
            sum=sum+a[i]*b[j];
        }
        i--;
        j--;
    }
    cout<<sum<<endl;
    return 0;
}