PAT 甲级 1037 Magic Coupon (25 分) 好水的贪心-蒲公英云

PAT 甲级 1037 Magic Coupon (25 分) 好水的贪心

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:

Sample Output:

43
#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;
}