1037 Magic Coupon

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:

Sample Output:

浅析

送分题

Code

#include 
#include 
#include 
using namespace std;
int main() {
    int m, n, ans = 0, p = 0, q = 0;
    scanf("%d", &m);
    int a1[m];
    for(int i = 0; i < m; i++)
        scanf("%d", &a1[i]);
    scanf("%d", &n);
    int a2[n];
    for(int i = 0; i < n; i++)
        scanf("%d", &a2[i]);
    sort(a1,a1+m);
    sort(a2,a2+n);
    while(p < m && q < n && a1[p] < 0 && a2[q] < 0) {
        ans += a1[p] * a2[q];
        p++; q++;
    }
    p = m - 1, q = n - 1;
    while(p >= 0 && q >= 0 && a1[p] > 0 && a2[q] > 0) {
        ans += a1[p] * a2[q];
        p--; q--;
    }
    printf("%d", ans);
    return 0;
}