Pat(Advanced Level)Practice--1037(Magic Coupon)

Pat 1037コード
タイトルの説明:
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

ACコード:

#include<cstdio>
#include<vector>
#include<algorithm>
#include<functional>

using namespace std;

int main(int argc,char *argv[])
{
	int Nc,Np;
	vector<int> P_Nc,N_Nc;//stand for positive and negative number of Nc
	vector<int> P_Np,N_Np;//the same for Np
	int i,j;
	scanf("%d",&Nc);
	for(i=0;i<Nc;i++)
	{
		int temp;
		scanf("%d",&temp);
		if(temp>0)
			P_Nc.push_back(temp);
		else if(temp<0)
			N_Nc.push_back(temp);
	}
	scanf("%d",&Np);
	for(j=0;j<Np;j++)
	{
		int temp;
		scanf("%d",&temp);
		if(temp>0)
			P_Np.push_back(temp);
		else if(temp<0)
			N_Np.push_back(temp);
	}
	sort(P_Nc.begin(),P_Nc.end(),greater<int>());
	sort(N_Nc.begin(),N_Nc.end());
	sort(P_Np.begin(),P_Np.end(),greater<int>());
	sort(N_Np.begin(),N_Np.end());
	long long max=0;
	i=j=0;
	while(i<P_Nc.size()&&j<P_Np.size())
	{
		max+=P_Nc[i]*P_Np[j];
		i++;
		j++;
	}
	i=j=0;
	while(i<N_Nc.size()&&j<N_Np.size())
	{
		max+=N_Nc[i]*N_Np[j];
		i++;
		j++;
	}
	printf("%lld
",max);

	return 0;
}

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