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

感想：这道题先排序再在数组两边分别计算