甲级PAT 1070 Mooncake （排序+贪心）
1070 Mooncake （25 分）
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 (≤1000), the number of different kinds of mooncakes, and D (≤500 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题目要求 ：
给定所有种类的月饼的库存量，总售价，以及市场最大的销售量。在不超过市场最大销售量的情况下如何售卖获取最大利润，结果保留两位小数。
解题思路：
算出每类月饼的单价，按从大到小排序，每次选取单价最大的直到达到最大销售量
注意：
月饼的库存量以及最大销售量一定要用浮点数存储。虽然题目给的是整数，但是如果用整数存储第三个测试点始终通不过。
完整代码：
#include
using namespace std;
struct Mooncake{
double amount;
double price;
double perpro;
}mooncake[1001];
bool comp(Mooncake m1,Mooncake m2){
return m1.perpro>m2.perpro;
}
int main(){
int N,i;
double profit=0.0,D;
scanf("%d%lf",&N,&D);
for(i=0;i 代码简洁
2019.2.8
#include
using namespace std;
struct Cake{
double price,amount,perprice;
};
bool com(Cake c1,Cake c2){
return c1.perprice>c2.perprice;
}
int main(){
int n,i;
double d,total=0,temp=0;
Cake cake[1010];
scanf("%d %lf",&n,&d);
for(i=0;i=d) break;
}
printf("%.2f",total);
return 0;
}