pat 1037

1037. Magic Coupon (25)

时间限制

100 ms

内存限制

32000 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:

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

Sample Output:

43

//刚开始看这题还以为要用dp什么的呢，没想到这么水啊。
#include <stdio.h>
#include <algorithm>
using namespace std;

int nc, np;
int cp[100010], pd[100010];

bool cmp(int a, int b)
{
	return a > b;
}
int main()
{
	int i, j, l;

	while(scanf("%d", &nc) != EOF)
	{
		for(i = 0; i < nc; i++)
			scanf("%d", &cp[i]);

		sort(cp, cp + nc, cmp);

		scanf("%d", &np);
		for(i = 0; i < np; i++)
			scanf("%d", &pd[i]);

		sort(pd, pd + np, cmp);
		
		int sum = 0;

		for(i = 0; i < np && i < nc; i++)
			if(cp[i] > 0 && pd[i] > 0)
				sum += cp[i] * pd[i];
			else
				break;

		for(l = nc - 1, j = np - 1; l >= 0 && j >= 0; l--, j--)
			if(cp[l] < 0 && pd[j] < 0)
				sum += cp[l] * pd[j];
			else
				break;

			printf("%d\n", sum);
		
	}
	return 0;
}