PAT A 1037. Magic Coupon (25)

题目

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

 

即求最大的couduct和coupons乘积乘积和，

排序，由大到小取正数相乘，由绝对值大到小取负数相乘，累加即可。

 

代码：

#include <iostream>#include <deque>#include <algorithm>using namespace std;int main(){int nc,np;long long money=0;int i,temp;deque<int> pos_c,neg_c,pos_p,neg_p;//存储正负coupon，正负productcin>>nc;//输入for(i=0;i<nc;i++){scanf("%d",&temp);if(temp>0)pos_c.push_back(temp);else if(temp<0)neg_c.push_back(temp);}cin>>np;for(i=0;i<np;i++){scanf("%d",&temp);if(temp>0)pos_p.push_back(temp);else if(temp<0)neg_p.push_back(temp);}sort(pos_c.begin(),pos_c.end());//排序sort(neg_c.begin(),neg_c.end());sort(pos_p.begin(),pos_p.end());sort(neg_p.begin(),neg_p.end());while(!pos_c.empty()&&!pos_p.empty())//加正的相乘{money+=pos_c.back()*pos_p.back();pos_c.pop_back();pos_p.pop_back();}while(!neg_c.empty()&&!neg_p.empty())//加负的相乘{money+=neg_c.front()*neg_p.front();neg_c.pop_front();neg_p.pop_front();}cout<<money;return 0;}