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≤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:

Sample Output:

题目大意

给你俩个数列A与B，你要从A与B当中，选取相同数量的元素然后⼀⼀相乘，求这样能得到

的乘积之和最⼤是多少

思路

将俩个数列排序后，最大的正数依次相乘，最大的负数依次相乘即可

C/C++


#include
using namespace std;
typedef long long LL;
int main()
{
    LL N,M,sum=0,num;
    vector N1,N2,M1,M2;
    cin >> N;
    while (N--){
        cin >> num;
        if(num>0) N1.push_back(num);
        else N2.push_back(num);
    }
    cin >> M;
    while (M--){
        cin >> num;
        if(num>0) M1.push_back(num);
        else M2.push_back(num);
    }
    sort(N1.begin(),N1.end());
    sort(N2.begin(),N2.end());
    sort(M1.begin(),M1.end());
    sort(M2.begin(),M2.end());
 
    while (!N1.empty() && !M1.empty()){
        sum += N1.back() * M1.back();
        N1.pop_back();
        M1.pop_back();
    }
 
    LL len = min(N2.size(),M2.size());
    for(int z=0;z
    cout << sum << endl;
    return 0;
}

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原文地址：https://blog.csdn.net/daybreak_alonely/article/details/127758773