Mooncake is a Chinese bakery product traditionally eaten during the Mid-Autumn Festival. Many types of fillings and crusts can be found in traditional mooncakes according to the region's culture. Now given the inventory amounts and the prices of all kinds of the mooncakes, together with the maximum total demand of the market, you are supposed to tell the maximum profit that can be made.

Note: partial inventory storage can be taken. The sample shows the following situation: given three kinds of mooncakes with inventory amounts being 180, 150, and 100 thousand tons, and the prices being 7.5, 7.2, and 4.5 billion yuans. If the market demand can be at most 200 thousand tons, the best we can do is to sell 150 thousand tons of the second kind of mooncake, and 50 thousand tons of the third kind. Hence the total profit is 7.2 + 4.5/2 = 9.45 (billion yuans).

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers

Output Specification:

For each test case, print the maximum profit (in billion yuans) in one line, accurate up to 2 decimal places.

Sample Input:

3 200
180 150 100
7.5 7.2 4.5

Sample Output:

9.45

algorithm design

This question is a classic application of greedy algorithm. Of moon cake can be given in descending order according to unit price, the unit price will sell some more and more moon cake as far as possible, be able to get the maximum benefit.

important point

Inventory given the title as well as a variety of moon cake did not specify the unit price is a positive integer, so it is necessary to store a double, otherwise there is a wrong answer to a test point will happen

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 using namespace std;
 5 struct Node
 6 {
 7     double qty, prof, pric;
 8 }node;
 9 int N, D;
10 int main()
11 {
12     cin >> N >> D;
13     vector<Node>Cake;
14     double res = 0.0;
15     for (int i = 0; i < N; ++i)
16     {
17         cin >> node.qty;
18         Cake.push_back(node);
19     }
20     for (int i = 0; i < N; ++i)
21     {
22         cin >> Cake[i].prof;
23         Cake[i].pric = Cake[i].prof / Cake[i].qty;
24     }
25     sort(Cake.begin(), Cake.end(), [](Node a, Node b) {return a.pric > b.pric; });
26     for (int i = 0; i < N && D>0; ++i)
27     {
28         int buy = Cake[i].qty < D ? Cake[i].qty : D;
29         D -= buy;
30         res += buy * Cake[i].pric;
31     }
32     printf("%.2f\n", res);
33     return 0;
34 }

PAT Grade --A1070 Mooncake