PAT1037. Magic Coupon

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思路：也算是贪心吧， 对两组数据分别进行排序 正对正 负对负

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 #define MAX 100010
 5 long long  coupon[MAX];
 6 long long  product[MAX];
 7 bool cmp(long long  a,long long  b)
 8 {
 9     return a>b;
10 }
11 int main(int argc, char *argv[])
12 {
13     int Np,Nc;
14     scanf("%d",&Nc);
15     for(int i=0;i<Nc;i++)
16        scanf("%lld",&coupon[i]);
17     scanf("%d",&Np);
18     for(int i=0;i<Np;i++)
19        scanf("%lld",&product[i]);
20     sort(coupon,coupon+Nc,cmp);
21     sort(product,product+Np,cmp);
22     long long get=0;
23     int j=0;
24     //正数
25     for(int i=0;i<Nc;i++)
26     {
27         if(coupon[i]>0&&product[j]>0)
28         {
29           get+=coupon[i]*product[j++];
30         }
31     }
32     //负数
33     j=Np-1;
34     for(int i=Nc-1;i>=0;i--)
35     {
36         if(coupon[i]<0&&product[j]<0)
37         {
38             get+=coupon[i]*product[j--];
39         }
40     }
41     printf("%lld\n",get);
42
43     return 0;
44 }