PAT 1037 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 NCcoupon 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
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
bool Compare(int x,int y)
{
return x>y;
}
int main()
{
int N,num,index;
vector<int> x;
vector<int> y;
vector<int> u;
vector<int> v;
scanf("%d",&N);
for(int i=0;i<N;i++)
{
scanf("%d",&num);
if(num>0)
x.push_back(num);
else
y.push_back(fabs(num));
}
sort(x.begin(),x.end(),Compare);
sort(y.begin(),y.end(),Compare);
scanf("%d",&N);
for(int i=0;i<N;i++)
{
scanf("%d",&num);
if(num>0)
u.push_back(num);
else
v.push_back(fabs(num));
}
sort(u.begin(),u.end(),Compare);
sort(v.begin(),v.end(),Compare);
LL ans=0;
int lx=x.size();
int ly=y.size();
int lu=u.size();
int lv=v.size();
index=0;
while(index<lx&&index<lu)
{
ans+=x[index]*u[index];
index+=1;
}
index=0;
while(index<ly&&index<lv)
{
ans+=y[index]*v[index];
index+=1;
}
printf("%lld",ans);
return 0;
}