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 -3Sample Output:
43
思路
贪心算法，
先排序，然后正数与正数相乘，负数与负数相乘就能得到最大值。
代码
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
int NC,NP;
cin >> NC;
vector<ll> coupons(NC);
for(int i = 0;i < NC;i++)
{
cin >> coupons[i];
}
cin >> NP;
vector<ll> products(NP);
for(int i = 0;i < NP;i++)
{
cin >> products[i];
}
sort(coupons.begin(),coupons.end());
sort(products.begin(),products.end());
int sum = 0;
int i = 0,j = 0,lc = coupons.size() - 1,lp = products.size() - 1;
for(;i <= lc && j <= lp;i++,j++)
{
if( coupons[i] <= 0 && products[j] <= 0 )
{
sum += coupons[i] * products[j];
}
}
for(int u = lc,v = lp;u >=0 && v >=0;u--,v--)
{
if( coupons[u] > 0 && products[v] > 0 )
{
sum += coupons[u] *products[v];
}
}
cout << sum << endl;
}