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1037 Magic Coupon (25 分)

时间:2019-01-26 20:24:53      阅读:156      评论:0      收藏:0      [点我收藏+]

标签:ber   sel   bre   values   cts   back   eset   font   more   

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 ? }, and a set of product values { 7 6 ? ? } (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, and it is guaranteed that all the numbers will not exceed 2?30??.

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

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<string.h>
#include<cstdio>
#include<cmath>
using namespace std;







int main()
{
    int nc;
    scanf("%d",&nc);
    vector<int> coupons(nc);
    for(int i=0; i<nc; i++)
    {
        scanf("%d",&coupons[i]);
    }
    int np;
    scanf("%d",&np);
    vector<int>products(np);
    //products.resize(np);
    for(int i=0; i<np; i++)
    {
        scanf("%d",&products[i]);
    }
    sort(coupons.begin(),coupons.end(),greater<int>());

    sort(products.begin(),products.end(),greater<int>());
    int sum=0;
    for(int i=0; i<nc&&i<np; i++)
    {
       // cout<<coupons[i];
        if(coupons[i]>0&&products[i]>0)
            sum+=coupons[i]*products[i];
        else
            break;
    }
    reverse(coupons.begin(),coupons.end());
    reverse(products.begin(),products.end());
    for(int i=0; i<nc&&i<np; i++)
    {
        if(coupons[i]<0&&products[i]<0)
            sum+=coupons[i]*products[i];
        else
            break;
    }
    cout<<sum<<endl;
    return 0;
}

 

1037 Magic Coupon (25 分)

标签:ber   sel   bre   values   cts   back   eset   font   more   

原文地址:https://www.cnblogs.com/zhanghaijie/p/10324756.html

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