1037 Magic Coupon

三块不一样的石头

于 2022-11-08 22:54:52 发布

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分类专栏： PAT (Advanced Level) Practice 文章标签：算法 PTA 贪心算法

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本文介绍了一个关于如何在火星上的魔法商店通过合理搭配使用魔法券和商品来最大化收益的问题。通过对两个列表进行排序并匹配最大正数和最小负数的方式，可以实现利益的最大化。

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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<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
    LL N,M,sum=0,num;
    vector<LL> 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<len;z++) sum += N2[z] * M2[z];
    cout << sum << endl;
    return 0;
}