PAT A组1037题(Magic Coupon)

题目描述

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.

（大意是两个整数集合，不重复分别选择相同个数的数一对一相乘，求能得到乘积之和的最大值。）

思路&代码实现

将两个集合都从小到大排序，然后分成负数乘积之和以及正数乘积之和。由于已经排序，乘积最大的情况就是各自绝对值最大的情况。

AC代码

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int Nc, Np;
    long long ans = 0;
    cin >> Nc;
    int coupon[Nc];
    for (int i = 0; i < Nc; i++)
        cin >> coupon[i];
    cin >> Np;
    int values[Np];
    for (int i = 0; i < Np; i++)
        cin >> values[i];
    sort(coupon, coupon + Nc);
    sort(values, values + Np);
    int k1 = 0, k2 = Np - 1;
    while (k1 < Nc && k1 < Np)
    {
        if (coupon[k1] < 0 && values[k1] < 0)
            ans += coupon[k1] * values[k1];
        else
            break;
        k1++;
    }
    k1 = Nc - 1;
    while (k1 >= 0 && k2 >= 0)
    {
        if (coupon[k1] > 0 && values[k2] > 0)
            ans += coupon[k1] * values[k2];
        else
            break;
        k1--;
        k2--;
    }
    cout << ans << endl;
    return 0;
}