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 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≤N C ,N P ≤10 5 , 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 代码如下：
#include <iostream> #include <algorithm> #include <vector> #include <string> #include <math.h> using namespace std; const int maxn=100005; bool cmp(int a,int b){ return a>b; } int main(){ int nc,np; vector<int> acp,acn,app,apn;//acp表示正数的coupen acn表示负数的coupen app表示正数的product cin>>nc; int num; for(int i = 0;i < nc;i++){ cin>>num; if(num>0){ //0不用考虑 acp.push_back(num); }else{ acn.push_back(num); } } cin>>np; for(int i = 0;i < np;i++){ cin>>num; if(num>0){ app.push_back(num); }else{ apn.push_back(num); } } sort(acp.begin(),acp.end(),cmp);//正数从大到小 sort(app.begin(),app.end(),cmp); sort(acn.begin(),acn.end());//负数从小到大 sort(apn.begin(),apn.end()); int money=0; vector<int>::iterator it1,it2; for(it1=acp.begin(),it2=app.begin();it1!=acp.end()&&it2!=app.end();it1++,it2++){ money+=(*it1)*(*it2); } for(it1=acn.begin(),it2=apn.begin();it1!=acn.end()&&it2!=apn.end();it1++,it2++){ money+=(*it1)*(*it2); } cout<<money<<endl; return 0; }