PAT_A1037#Magic Coupon

Source:

PAT A1037 Magic Coupon (25 分)

Description:

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 − }, and a set of product values { 7 6 − − } (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with Nbeing 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 N​C​​, followed by a line with N​C​​ coupon integers. Then the next line contains the number of products N​P​​, followed by a line with N​P​​ product values. Here 1, and it is guaranteed that all the numbers will not exceed 2​30​​.

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

Keys:

• 贪心

Code:

 /*
 Data: 2019-07-23 19:10:20
 Problem: PAT_A1037#Magic Coupon
 AC: 11:55

 题目大意：
 集合A和集合B，求集合A中各元素与集合B中各元素乘积之和的最大值
 */
 #include<cstdio>
 #include<algorithm>
 #include<functional>
 using namespace std;
 const int M=1e5+;
 int c[M],p[M];

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int nc,np,pos,ans=;
     scanf("%d", &nc);
     for(int i=; i<nc; i++)
         scanf("%d", &c[i]);
     scanf("%d", &np);
     for(int i=; i<np; i++)
         scanf("%d", &p[i]);
     sort(c,c+nc,greater<int>() );
     sort(p,p+np,greater<int>() );
     pos=;
     while(pos<nc && pos<np && p[pos]> && c[pos]>)
         ans += p[pos]*c[pos++];
     sort(c,c+nc,less<int>() );
     sort(p,p+np,less<int>() );
     pos=;
     while(pos<nc && pos<np && p[pos]< && c[pos]<)
         ans += p[pos]*c[pos++];
     printf("%d", ans);

     return ;
 }