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.

41 2 4 -147 6 -2 -3Sample Output:
43
#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <map>#include <string>#define Max 100000using namespace std;bool cmp(int a,int b){	return a>b;}int main(){	int n,m;	int S[Max],K[Max];	scanf("%d",&n);	for(int i=0;i<n;i++)	{		scanf("%d",&S[i]);	}	scanf("%d",&m);	for(int i=0;i<m;i++)	{		scanf("%d",&K[i]);	}	sort(S,S+n,cmp);	sort(K,K+m,cmp);	int num=0,f;	for(int i=0;i<n&&i<m;i++)	{		if(S[i]>0&&K[i]>0)		{			num+=S[i]*K[i];					}		else 		{			f=i;			break;		}	}	for(int i=n-1;i>=f;i--)	{		for(int j=m-1;j>=f;j--)		{			m=m-1;			if(S[i]<0&&K[j]<0)			{				num+=S[i]*K[j];								break;			}		}	}	printf("%d/n",num);    system("pause");	return 0;}