1037. Magic Coupon (25)

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

#include<iostream>
#include<algorithm>
using namespace std;
int c[100005],p[100005];
int main(){
    freopen("in.txt","r",stdin);
    int nc,np;
    cin>>nc;
    for(int i=0;i<nc;i++)
        cin>>c[i];
    cin>>np;
    for(int i=0;i<np;i++)
        cin>>p[i];
    sort(c,c+nc);
    sort(p,p+np);
    int k1=0,k2=0,e1=nc-1,e2=np-1,ans=0;

    while(c[k1]<0&&p[k2]<0){
            ans+=c[k1]*p[k2];
            k1++,k2++;
            if(k1==nc||k2==np) break;
    }       
    k1=nc-1,k2=np-1;
    while(c[k1]>0&&p[k2]>0){
            ans+=c[k1]*p[k2];
            k1--,k2--;
            if(k1==-1||k2==-1) break;
    }
    cout<<ans<<endl;

    return 0;
}