1070 Mooncake （25 分）

最新推荐文章于 2020-03-01 09:50:14 发布

xiaoshengMr

最新推荐文章于 2020-03-01 09:50:14 发布

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分类专栏： PAT甲级(Advanced Level) 文章标签： PAT

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本文介绍了一种计算在给定市场总需求下，如何通过合理选择不同种类的月饼库存进行销售，以达到最大利润的方法。通过计算每种月饼的单位价格，采用贪心算法选取单价最高的月饼进行销售，直至满足市场需求。

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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

思路就是求出每中月饼的单价，简单模拟，贪心取单价最高的月饼，然后求出最后的总售价，注意数据类型，都为double

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1010;
struct moon{
   double sale;
   double sigprice;
   double allprice;
}mons[maxn];
bool cmp(moon m1,moon m2)
{
	return m1.sigprice>m2.sigprice;
}
int n;
double d;
int main()
{
	//freopen("C:\\Users\\Aixscode\\Desktop\\input.txt","r",stdin);
	cin>>n>>d;
	for(int i=0;i<n;i++)
	{
		cin>>mons[i].sale;
	}
	for(int i=0;i<n;i++)
	{
		cin>>mons[i].allprice;
	}
	for(int i=0;i<n;i++)
	{
		mons[i].sigprice=mons[i].allprice/mons[i].sale;
	}
	sort(mons,mons+n,cmp);
	double allprice=0;
	double tmp=d;
	for(int i=0;i<n;i++)
	{
		if(mons[i].sale>=tmp){
			allprice+=tmp*mons[i].sigprice;
			break;
		}
		else 
		{
		    tmp-=mons[i].sale;
			allprice+=mons[i].allprice;
		}
	}
	printf("%.2f\n",allprice);
    return 0;
}