点击链接PAT甲级-AC全解汇总

题目：
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
1
2
3
4

Sample Output:

43
1

题意：
求两组数各挑一个数相乘的和最大，每个数只能用一次，可以不用

我的思路：
把所有正数从大到小排，大的乘大的求和；
把所有负数从小到大排，小的乘小的求和；
剩下一正一负就算了，求和反而变小了。

我的代码：

#include
using namespace std;

int main(){
    vector<int>coupon_p,coupon_n;
    vector<int>product_p,product_n;
    int N;
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int t;
        cin>>t;
        if(t>0)coupon_p.push_back(t);
        else coupon_n.push_back(t);
    }
    cin>>N;
    for(int i=0;i<N;i++)
    {
        int t;
        cin>>t;
        if(t>0)product_p.push_back(t);
        else product_n.push_back(t);
    }
    sort(coupon_p.begin(),coupon_p.end(),greater<int>());
    sort(coupon_n.begin(),coupon_n.end());
    sort(product_p.begin(),product_p.end(),greater<int>());
    sort(product_n.begin(),product_n.end());

    int ans=0;
    for(int i=0;i<min(coupon_p.size(),product_p.size());i++)
    {
        ans+=coupon_p[i]*product_p[i];
    }
    for(int i=0;i<min(coupon_n.size(),product_n.size());i++)
    {
        ans+=coupon_n[i]*product_n[i];
    }
    cout<<ans<<endl;

    return 0;
}

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