1037 Magic Coupon（贪心，排序）-蒲公英云

1037 Magic Coupon（贪心，排序）

我就是我 2024-04-08 12:44 153阅读 0赞

1037 Magic Coupon（贪心，排序）

0、题目

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 N\*C, followed by a line with N**C coupon integers. Then the next line contains the number of products N**P, followed by a line with N**P product values. Here 1≤N**C,N**P*≤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:

1、大致题意

给出两个数字序列，从这两个序列中分别选取相同数量的元素进行一对一相乘，问能得到的乘积之和最大为多少

2、基本思路

把这两个序列在存储时，将正数和负数分开，并将正数从大到小排序，负数从小到大排序。将前面都是负数的数相乘求和，然后将后面都是正数的数相乘求和。

3、AC代码

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int n,m,tmp;
vector<int>az,af,bz,bf;
int main() {
    cin>>m;
    for(int i=0; i<m; i++) {
        cin>>tmp;
        if(tmp>0) {
            az.push_back(tmp);
        } else {
            af.push_back(tmp);
        }
    }
    sort(az.begin(),az.end(),greater<int>());
    sort(af.begin(),af.end(),less<int>());
    cin>>n;
    for(int i=0; i<n; i++) {
        cin>>tmp;
        if(tmp>0) {
            bz.push_back(tmp);
        } else {
            bf.push_back(tmp);
        }
    }
    sort(bz.begin(),bz.end(),greater<int>());
    sort(bf.begin(),bf.end(),less<int>());
    int ans=0;
    int size=min(az.size(),bz.size());
    for(int i=0; i<size; i++) {
        ans+=az[i]*bz[i];
    }
    size=min(af.size(),bf.size());
    for(int i=0; i<size; i++) {
        ans+=af[i]*bf[i];
    }
    cout<<ans;
    return 0;
}