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 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 <= 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:

4
1 2 4 -1
4
7 6 -2 -3
Sample Output:

分析:简单的贪心应用

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int n, m, total = 0;
    cin >> n;
    vector<int>a(n);
    for (int i = 0; i <n; i++)
       cin >> a[i];
    cin >> m;
    vector<int>b(m);
    for (int i = 0; i < m; i++)
        cin >> b[i];
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    for (int i = 0, j = 0; i < a.size() && j < b.size() && a[i] < 0 && b[j] < 0; i++, j++)
        total+= (a[i] * b[j]);
    for (int i = a.size()-1, j = b.size()-1; i >= 0 && j >= 0 && a[i] > 0 && b[j] > 0; i--, j--)
        total += (a[i] * b[j]);

    cout << total;

    return 0;
}

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