1037. Magic Coupon

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:

41 2 4 -147 6 -2 -3

Sample Output:

43

//有个bug，用scanf函数作为输入的时候，你会发现，sort排序，对于正数和负数的排序结果是分开的，以后用到sort的时候，得不到最终的结果，最好改用到cin进行输入，

//1037. Magic Coupon//你可以把cin，cout换成scanf，printf再跑一下试试。test sort()打印一下v1，v2中的值看看 #include<iostream>#include<vector>#include<algorithm>using namespace std;vector<long long> v1;vector<long long> v2;int main(){int n, m;cin>>n;for(int i = 0; i < n; i ++){long long tmp;cin>>tmp;v1.push_back(tmp);}cin>>m;if(n == 0 || m == 0){cout<<"0"<<endl;return 0;}for(int i = 0; i < m; i ++){long long tmp;cin>>tmp;v2.push_back(tmp);}sort(v1.begin(), v1.end());sort(v2.begin(), v2.end());//test sort()int start1 = 0, end1 = v1.size()-1;int start2 = 0, end2 = v2.size()-1;long long sum = 0;while(start1 <= end1 && start2 <= end2 && v1[start1] < 0 && v2[start2] < 0){sum += v1[start1]*v2[start2];start1 ++;start2 ++;}while(end1 >= 0 && end2 >= 0 && v1[end1] > 0 && v2[end2] > 0){sum += v1[end1]*v2[end2];end1 --;end2 --;}cout<<sum<<endl;return 0;}