乘积最大值（贪心思想）

问题

给出两个集合，各选出n个元素，求乘积最大值、
1037 Magic Coupon (25 分)
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
​
≤10
5
, and it is guaranteed that all the numbers will not exceed 2
30
.

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
结尾无空行
Sample Output:
43

思路

把两个集合的正负数分开，正数从大到小排序，负数从小到大排序。相应位置相乘即为最大值。

代码

#include
using namespace std;

vector<int> set1_pos,set2_pos,set1_neg,set2_neg;
bool cmp(int a,int b){
    return a>b;
}
int main(){
    int n1,n2,temp;
    cin>>n1;
    for (int i = 0; i < n1; ++i) {
        cin>>temp;
        if(temp>=0){
            set1_pos.push_back(temp);
        } else{
            set1_neg.push_back(temp);
        }
    }
    cin>>n2;
    for (int j = 0; j < n2; ++j) {
        cin>>temp;
        if(temp>=0){
            set2_pos.push_back(temp);
        } else{
            set2_neg.push_back(temp);
        }
    }
    int len1 = min(set1_pos.size(),set2_pos.size());
    int len2 = min(set1_neg.size(),set2_neg.size());
    sort(set1_pos.begin(),set1_pos.end(),cmp);
    sort(set2_pos.begin(),set2_pos.end(),cmp);
    sort(set1_neg.begin(),set1_neg.end());
    sort(set2_neg.begin(),set2_neg.end());
    int ans = 0;
    for (int k = 0; k < len1; ++k) {
        ans = ans + set1_pos[k]*set2_pos[k];
    }
    for (int l = 0; l < len2; ++l) {
        ans = ans + set1_neg[l]*set2_neg[l];
    }
    cout<<ans;
    return 0;
}

总结

该题的时间限制较宽，使用了四次sort函数。