Magic Coupon (25)

题目描述

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.

输入描述:

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.

输出描述:

For each test case, simply print in a line the maximum amount of money you can get back.

输入例子:

41 2 4 -147 6 -2 -3

输出例子:

43

我的代码：

#include<iostream>#include<algorithm>using namespace std;long long a[100001]={0},b[100001]={0},sum=0;int main(){    int n,m,i,j;    cin>>n;    for(i=0;i<n;i++) cin>>a[i];    sort(a,a+n);    cin>>m;    for(i=0;i<m;i++) cin>>b[i];    sort(b,b+m); i=0;    while(i<n && i<m && a[i]<0 && b[i]<0)    {        sum=sum+a[i]*b[i]; i++;    }    i=n-1,j=m-1;    while(i>=0 && j>=0 && a[i]>0 && b[j]>0)        sum=sum+a[i--]*b[j--];    cout<<sum<<endl;    return 0;}