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
IDEA
1.把coupon和product正数和负数分别分两组,正数组从大到小排序,负数组从小到大排序
正数组中,coupon大数*product大数 叠加
负数组中,coupon小数*product小数 叠加
二者和为结果
CODE
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#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int cmp1(int a,int b){
return a>b;
}
int cmp2(int a,int b){
return a<b;
}
int main(){
int nc,np;
vector<int> c_pos,c_neg;
vector<int> p_pos,p_neg;
cin>>nc;
for(int i=0;i<nc;i++){
int coupon;
cin>>coupon;
if(coupon>0){
c_pos.push_back(coupon);
}else if(coupon<0){
c_neg.push_back(coupon);
}
}
cin>>np;
for(int i=0;i<np;i++){
int product;
cin>>product;
if(product>0){
p_pos.push_back(product);
}else if(product<0){
p_neg.push_back(product);
}
}
sort(c_pos.begin(),c_pos.end(),cmp1);
sort(c_neg.begin(),c_neg.end(),cmp2);
sort(p_pos.begin(),p_pos.end(),cmp1);
sort(p_neg.begin(),p_neg.end(),cmp2);
int sum=0;
for(int i=0,j=0;i<c_pos.size()&&j<p_pos.size();i++,j++){
sum+=c_pos[i]*p_pos[j];
}
for(int i=0,j=0;i<c_neg.size()&&j<p_neg.size();i++,j++){
sum+=c_neg[i]*p_neg[j];
}
cout<<sum;
return 0;
}