PAT_A1037#Magic Coupon
Source:
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 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, 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
Keys:
Code:
1 /* 2 Data: 2019-07-23 19:10:20 3 Problem: PAT_A1037#Magic Coupon 4 AC: 11:55 5 6 题目大意： 7 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 8 */ 9 #include<cstdio> 10 #include<algorithm> 11 #include<functional> 12 using namespace std; 13 const int M=1e5+10; 14 int c[M],p[M]; 15 16 int main() 17 { 18 #ifdef ONLINE_JUDGE 19 #else 20 freopen("Test.txt", "r", stdin); 21 #endif 22 23 int nc,np,pos,ans=0; 24 scanf("%d", &nc); 25 for(int i=0; i<nc; i++) 26 scanf("%d", &c[i]); 27 scanf("%d", &np); 28 for(int i=0; i<np; i++) 29 scanf("%d", &p[i]); 30 sort(c,c+nc,greater<int>() ); 31 sort(p,p+np,greater<int>() ); 32 pos=0; 33 while(pos<nc && pos<np && p[pos]>0 && c[pos]>0) 34 ans += p[pos]*c[pos++]; 35 sort(c,c+nc,less<int>() ); 36 sort(p,p+np,less<int>() ); 37 pos=0; 38 while(pos<nc && pos<np && p[pos]<0 && c[pos]<0) 39 ans += p[pos]*c[pos++]; 40 printf("%d", ans); 41 42 return 0; 43 }
