PAT(甲级)1037

1037. Magic Coupon (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

41 2 4 -147 6 -2 -3

Sample Output:

43

#include <cstdio>#include <algorithm>#define SIZE 100005int Coupon[SIZE];int Product[SIZE];void swap(int &a,int &b){int tmp=a;a=b;b=tmp;}int main(){int NC,NP;scanf("%d",&NC);int i=0;for(i=0;i<NC;i++)    scanf("%d",&Coupon[i]);scanf("%d",&NP);for(i=0;i<NP;i++)    scanf("%d",&Product[i]);std::sort(Coupon,Coupon+NC);std::sort(Product,Product+NP);for(i=0;i<NC;i++){if(Coupon[i] > 0 )    break;}int cfirst=i;for(i=0;i<NP;i++)    if(Product[i] > 0)        break;int pfirst=i;int min = cfirst < pfirst ? cfirst:pfirst;long long res=0;for(i= 0 ;i<min;i++)    res +=Coupon[i]*Product[i];cfirst = NC-cfirst;pfirst = NP-pfirst;min = cfirst < pfirst ? cfirst:pfirst;for(i=0;i<min;i++)    res += Coupon[NC-1-i]*Product[NP-1-i];    printf("%lld\n",res);    return 0;}