A1037 Magic Coupon (25)

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:

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

Sample Output:

43

注意，nc,np的值未必相同

#include<cstdio>
#include<algorithm>

using namespace std;
int main(){
int nc,np,sum = 0;
scanf("%d",&nc);
int a[nc];
for(int i = 0;i < nc;i++)
scanf("%d",&a[i]);
scanf("%d",&np);
int b[np];
for(int i = 0;i < np;i++)
scanf("%d",&b[i]);
sort(a,a+nc);
sort(b,b+np);
int i = nc-1,j = np-1;
while(i>=0&&j>=0&&a[i]*b[j]>0&&a[i]>0)		//在总数相同及正负分布相同时停下来
sum += a[i--]*b[j--];
i = 0,j = 0;
while(i<nc&&j<np&&a[i]*b[j]>0&&a[i]<0)
sum += a[i++]*b[j++];
printf("%d\n",sum);
return 0;
}