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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
https://www.patest.cn/contests/pat-a-practise/1037
思路：
两类分别绝对值从大到小排序，然后正、负分别匹配。
CODE：
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp1(int a,int b)
{return a>b;
}
bool cmp2(int a,int b)
{return a<b;
}
int main()
{int n;cin>>n;int num1=0;int num2=0;int po[n];int ne[n];for (int i=0;i<n;i++){int t;cin>>t;if (t>0){po[num1]=t;num1++;}else{ne[num2]=t;num2++;}}int m;cin>>m;int mum1=0;int mum2=0;int po1[m];int ne1[m];for (int i=0;i<m;i++){int t;cin>>t;if(t>0){po1[mum1]=t;mum1++;}else{ne1[mum2]=t;mum2++;}}sort(po,po+num1,cmp1);sort(ne,ne+num2,cmp2);sort(po1,po1+mum1,cmp1);sort(ne1,ne1+mum2,cmp2);int re=0;for (int i=0;i<min(num1,mum1);i++)re+=po[i]*po1[i];for (int i=0;i<min(num2,mum2);i++)re+=ne[i]*ne1[i];cout<<re;return 0;
}
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