• 1037 Magic Coupon

    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:

    1. 4
    2. 1 2 4 -1
    3. 4
    4. 7 6 -2 -3

    Sample Output:

    43

    题目大意

    给你俩个数列A与B，你要从A与B当中，选取相同数量的元素然后⼀⼀相乘，求这样能得到

    的乘积之和最⼤是多少

    思路

    将俩个数列排序后，最大的正数依次相乘，最大的负数依次相乘即可

    C/C++ 

    1. #include
    2. using namespace std;
    3. typedef long long LL;
    4. int main()
    5. {
    6.     LL N,M,sum=0,num;
    7.     vector N1,N2,M1,M2;
    8.     cin >> N;
    9.     while (N--){
    10.         cin >> num;
    11.         if(num>0) N1.push_back(num);
    12.         else N2.push_back(num);
    13.     }
    14.     cin >> M;
    15.     while (M--){
    16.         cin >> num;
    17.         if(num>0) M1.push_back(num);
    18.         else M2.push_back(num);
    19.     }
    20.     sort(N1.begin(),N1.end());
    21.     sort(N2.begin(),N2.end());
    22.     sort(M1.begin(),M1.end());
    23.     sort(M2.begin(),M2.end());
    24.  
    25.     while (!N1.empty() && !M1.empty()){
    26.         sum += N1.back() * M1.back();
    27.         N1.pop_back();
    28.         M1.pop_back();
    29.     }
    30.  
    31.     LL len = min(N2.size(),M2.size());
    32.     for(int z=0;z
    33.     cout << sum << endl;
    34.     return 0;
    35. }

     

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  • 原文地址：https://blog.csdn.net/daybreak_alonely/article/details/127758773