PAT甲级——A1070 Mooncake

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 N (≤), the number of different kinds of mooncakes, and D (≤ thousand tons), the maximum total demand of the market. Then the second line gives the positive inventory amounts (in thousand tons), and the third line gives the positive prices (in billion yuans) of N kinds of mooncakes. All the numbers in a line are separated by a space.

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

算法设计

这道题是贪心算法的经典应用。可对给出的月饼按照单价从大到小进行排序，将单价多的月饼尽可能地卖多一些，便能得到最大的收益。

注意点

题目给定的各种月饼的库存量以及单价并未说明是正整数，所以需要用double来储存，否则有一个测试点将会出现答案错误的情况

 #include <iostream>
 #include <vector>
 #include <algorithm>
 using namespace std;
 struct Node
 {
     double qty, prof, pric;
 }node;
 int N, D;
 int main()
 {
     cin >> N >> D;
     vector<Node>Cake;
     double res = 0.0;
     for (int i = ; i < N; ++i)
     {
         cin >> node.qty;
         Cake.push_back(node);
     }
     for (int i = ; i < N; ++i)
     {
         cin >> Cake[i].prof;
         Cake[i].pric = Cake[i].prof / Cake[i].qty;
     }
     sort(Cake.begin(), Cake.end(), [](Node a, Node b) {return a.pric > b.pric; });
     for (int i = ; i < N && D>; ++i)
     {
         int buy = Cake[i].qty < D ? Cake[i].qty : D;
         D -= buy;
         res += buy * Cake[i].pric;
     }
     printf("%.2f\n", res);
     return ;
 }