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:

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

Sample Output:

43

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int cmp(long long a, long long b)
{
return a > b;
}
int main()
{
int NC, NP;
scanf("%d", &NC);
vector<long long> negativecoupon, negativeproduct,
positivecoupon, positiveproduct;
long long temp;
for (int i = 0; i < NC; i++)
{
scanf("%lld", &temp);
if (temp >= 0)
positivecoupon.push_back(temp);
else
negativecoupon.push_back(temp);
}
scanf("%d", &NP);
for (int i = 0; i < NP; i++)
{
scanf("%lld", &temp);
if (temp >= 0)
4000

positiveproduct.push_back(temp);
else
negativeproduct.push_back(temp);
}
sort(positivecoupon.begin(), positivecoupon.end(), cmp);
sort(positiveproduct.begin(), positiveproduct.end(), cmp);
sort(negativecoupon.begin(), negativecoupon.end());
sort(negativeproduct.begin(), negativeproduct.end());
long long sum = 0;
for (int i = 0; i < positivecoupon.size() && i < positiveproduct.size(); i++)
{
sum += positivecoupon[i] * positiveproduct[i];
}
for (int i = 0; i < negativecoupon.size() && i < negativeproduct.size(); i++)
{
sum += negativecoupon[i] *negativeproduct[i];
}
printf("%lld", sum);
return 0;
}