1037. Magic Coupon (25)

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<cstdio>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
	int nc,np;
	scanf("%d",&nc);
	long long value;
	vector<long long> nc1;
	vector<long long> nc2;
	vector<long long> np1;
	vector<long long> np2;
	for(int i=0;i<nc;++i){
		scanf("%lld",&value);
		if(value>=0){
			nc1.push_back(value);
		}else{
			nc2.push_back(value);
		}
	}
	scanf("%d",&np);
	for(int i=0;i<np;++i){
		scanf("%lld",&value);
		if(value>0){
			np1.push_back(value);
		}else{
			np2.push_back(value);
		}
	}
	sort(nc1.begin(),nc1.end());
	sort(nc2.begin(),nc2.end());
	sort(np1.begin(),np1.end());
	sort(np2.begin(),np2.end());
	long long sum=0;
	if(nc1.size()&&np1.size()){
			for(int i=nc1.size()-1,j=np1.size()-1;i>=0&&j>=0;--i,--j){
				sum+=nc1[i]*np1[j];
			}
	}
	if(nc2.size()&&np2.size()){
		for(int i=0,j=0;i<nc2.size()&&j<np2.size();++i,++j){
			sum+=nc2[i]*np2[j];
		}
	}
	printf("%lld",sum);
	return 0;
}