1037 Magic Coupon（贪心，排序）

1037 Magic Coupon（贪心，排序）

0、题目

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≤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
1
2
3
4

Sample Output:

43
1

1、大致题意

给出两个数字序列，从这两个序列中分别选取相同数量的元素进行一对一相乘，问能得到的乘积之和最大为多少

2、基本思路

把这两个序列在存储时，将正数和负数分开，并将正数从大到小排序，负数从小到大排序。将前面都是负数的数相乘求和，然后将后面都是正数的数相乘求和。

3、AC代码

#include
#include
#include
using namespace std;
int n,m,tmp;
vector<int>az,af,bz,bf;

int main() {
	cin>>m;
	for(int i=0; i<m; i++) {
		cin>>tmp;
		if(tmp>0) {
			az.push_back(tmp);
		} else {
			af.push_back(tmp);
		}
	}
	sort(az.begin(),az.end(),greater<int>());
	sort(af.begin(),af.end(),less<int>());
	cin>>n;
	for(int i=0; i<n; i++) {
		cin>>tmp;
		if(tmp>0) {
			bz.push_back(tmp);
		} else {
			bf.push_back(tmp);
		}
	}
	sort(bz.begin(),bz.end(),greater<int>());
	sort(bf.begin(),bf.end(),less<int>());
	int ans=0;
	int size=min(az.size(),bz.size());
	for(int i=0; i<size; i++) {
		ans+=az[i]*bz[i];
	}
	size=min(af.size(),bf.size());
	for(int i=0; i<size; i++) {
		ans+=af[i]*bf[i];
	}
	cout<<ans;
	return 0;
}
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