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
代码如下
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int nc,np,ans = 0;
cin>>nc;
vector<int> v1(nc);
for(int i = 0; i < nc; i++){
scanf("%d", &v1[i]);
}
cin>>np;
vector<int> v2(np);
for(int i = 0; i < np; i++){
scanf("%d", &v2[i]);
}
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
for(int i = v1.size()-1,j = v2.size()-1; i>=0&&j>=0;i--,j--){
if(v1[i]>0&&v2[j]>0){
ans += v1[i]*v2[j];
}else{
break;
}
}
for(int i = 0,j = 0; i<v1.size()&&j<v2.size();i++,j++){
if(v1[i]<0&&v2[j]<0){
ans += v1[i]*v2[j];
}else{
break;
}
}
cout<<ans<<endl;
return 0;
}