PAT 甲级 1037 Magic Coupon (25 分) 好水的贪心

素颜马尾好姑娘i 2024-02-19 21:53 7阅读 0赞

## 1037 Magic Coupon (25 分) ##

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: ####

4
    1 2 4 -1
    4
    7 6 -2 -3

#### Sample Output: ####

#include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<vector>
    #include<cmath>
    using namespace std;
    int cmp(int a,int b)
    {
    	return a>b;
    }
    int main()
    {
    	int n,m,i,j,t,sum=0;
    	vector<int> zc,fc,zp,fp;
    	cin>>n;
    	for(i=0;i<n;i++)
    	{
    		cin>>t;
    		if(t>0)
    		zc.push_back(t);
    		else
    		fc.push_back(abs(t));
    	}
    	cin>>m;
    	for(i=0;i<m;i++)
    	{
    		cin>>t;
    		if(t>0)
    		zp.push_back(t);
    		else
    		fp.push_back(abs(t));
    	}
    	sort(zc.begin(),zc.end(),cmp);
    	sort(fc.begin(),fc.end(),cmp);
    	sort(zp.begin(),zp.end(),cmp);
    	sort(fp.begin(),fp.end(),cmp);
    	for(i=0;i<zc.size()&&i<zp.size();i++)
    	{
    		sum+=zc[i]*zp[i];
    	}
    	for(i=0;i<fc.size()&&i<fp.size();i++)
    	{
    		sum+=fc[i]*fp[i];
    	}
    	cout<<sum<<endl;
    	return 0;
    }