PAT甲级1037

Posted 2022-08-30 wx630c98f24f6b8

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 Ns!

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)
      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;