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 − }, and a set of product values { 7 6 − − } (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, 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
题意：
给出两个集合，从这两个集合里面选出数量相同的元素进行一对一相乘，求能够得到的最大乘积之和。
题解：
对每个集合，将正数和负数分开考虑，将每个集合里的整数从大到小排序；将每个集合里的负数从小到大排序，然后同位置的正数与正数相乘，负数与负数相乘。
注意点：
输入为0的不要管，直接忽略，否则测试点1过不去
AC代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll na[];
ll pa[];
ll nb[];
ll pb[];
int kna=,kpa=,knb=,kpb=;
int n1,n2;
bool cmp(ll x, ll y){
return x>y;
}
ll s=;
int main(){
cin>>n1;
ll x;
for(int i=;i<=n1;i++){
cin>>x;
if(x>){
pa[++kpa]=x;
}else if(x<){//=0的不要管
na[++kna]=x;
}
}
cin>>n2;
for(int i=;i<=n2;i++){
cin>>x;
if(x>){
pb[++kpb]=x;
}else if(x<){//=0的不要管
nb[++knb]=x;
}
}
sort(pa+,pa++kpa,cmp);
sort(pb+,pb++kpb,cmp);
sort(na+,na++kna);
sort(nb+,nb++knb);
int min_l=min(kpa,kpb);
for(int i=;i<=min_l;i++){
s+=pa[i]*pb[i];
}
min_l=min(kna,knb);
for(int i=;i<=min_l;i++){
s+=na[i]*nb[i];
}
cout<<s<<endl;
return ;
}