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:

41 2 4 -147 6 -2 -3

Sample Output:

43

#include <iostream>#include <vector>#include <algorithm>using namespace std;typedef long long ll;bool cmp(const ll &a, const ll &b){    return a > b;}int main(){    int n, m;    ll temp;    vector<ll> v1, v2;    cin >> n;    for(int i=0; i<n; i++)    {        cin >> temp;        v1.push_back(temp);    }    cin >> m;    for(int i=0; i<m; i++)    {        cin >> temp;        v2.push_back(temp);    }    sort(v1.begin(), v1.end(), cmp);    sort(v2.begin(), v2.end(), cmp);    int i, j, k;    ll total = 0;    for(i=0; v1[i]>0 && v2[i]>0; i++)    {        total += v1[i]*v2[i];    }    for(j=n-1, k=m-1; v1[j]<0 && v2[k]<0; j--, k--)    {        total += v1[j]*v2[k];    }    cout << total << endl;    return 0;}