题目如下：
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 4.5Sample Output:
9.45
题目要求按照利益最大化的原则出售月饼来满足市场需求，属于基本贪心问题。
应该按照公斤单价的降序排序，然后从单价最高的开始处理，直到处理完所有种类的月饼或者市场需求已经被满足。
需要注意的是，存货应该也用double存储，否则会有个case报错，可能是精度问题。
代码如下：
#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
struct Cake{
double inventory;
double price;
double average;
bool operator < (const Cake c) const{
return average > c.average;
}
};
int main()
{
int N,D;
cin >> N >> D;
vector<Cake> cakes(N);
for(int i = 0; i < N; i++){
scanf("%lf",&cakes[i].inventory);
}
for(int i = 0; i < N; i++){
scanf("%lf",&cakes[i].price);
cakes[i].average = cakes[i].price / cakes[i].inventory;
}
sort(cakes.begin(),cakes.end());
int cur = 0;
double sum = 0;
while(D && cur < N){
if(cakes[cur].inventory >= D){
sum += cakes[cur].average * D;
break;
}else{
sum += cakes[cur].price;
D -= cakes[cur].inventory;
cur++;
}
}
printf("%0.2lf\n",sum);
return 0;
}
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原文地址：http://blog.csdn.net/xyt8023y/article/details/47274973
