1037. Magic Coupon (25)-PAT甲级真题

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 M7) to get M28 back; coupon 2 to product 2 to get M12 back; and coupon 4 to product 4 to get M3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M12 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

　　题目大意：就是说有两个数组，分别从两个里面抽出一个相乘，执行n次，使得乘机总和最大。
　　分析:因为反正都是两个数组分别一个个取，那么久挨个判断，先从小到大排序，那么就挨个让最小的复数相乘，一直到没有负数可以匹配了，接着让最大的数互相乘，直到没有正数了，很容易啊。。。。。

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int main ()
{
    int m , n , result = 0 , p = 0 , q = 0;
    scanf_s ("%d" , &m);
    vector<int> v1 (m);
    for (int i = 0; i < m; i++)
        scanf_s ("%d" , &v1[i]);
    scanf_s ("%d" , &n);
    vector<int> v2 (n);
    for (int i = 0; i < n; i++)
        scanf_s ("%d" , &v2[i]);
    sort (v1.begin () , v1.end ());
    sort (v2.begin () , v2.end ());
    while (p < m && q < n && v1[p] < 0 && v2[q] < 0)
    {
        result += v1[p] * v2[q];
        p++; q++;
    }
    p = m - 1 , q = n - 1;
    while (p >= 0 && q >= 0 && v1[p] > 0 && v2[q] > 0)
    {
        result += v1[p] * v2[q];
        p--; q--;
    }
    printf ("%d" , result);
    return 0;
}