zoukankan      html  css  js  c++  java
  • 1070. Mooncake (25)

    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

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    struct Node{
    	double perprice;
    	double amount;
    	double money;
    };
    bool cmp(Node a, Node b){
    	return a.perprice>b.perprice;
    }
    int main(){
    	int i,j;
    	int N,D;
    	scanf("%d%d",&N,&D);
    	vector<Node>node;
    	node.resize(N);
    	for(i=0;i<N;i++){
    		scanf("%lf",&node[i].amount);
    	}
    	double tmp;
    	for(i=0;i<N;i++){
    		scanf("%lf",&tmp);
    		node[i].money=tmp;
    		node[i].perprice=tmp*1.0/node[i].amount;
    	}
    	sort(node.begin(),node.end(),cmp);
    	double sum=0;
    	i=0;
    	while(0<D){
    		if(D>=node[i].amount){
    			D-=node[i].amount;
    			sum+=node[i].money;
    		}else {
    			sum+=node[i].money*D*1.0/node[i].amount;
    			D=0;
    		}
    		i++;
    		if(i>=N)break;
    	}
    	printf("%.2lf
    ",sum);
    	return 0;
    }
    

      




  • 相关阅读:
    sql声明变量,及if -else语句、while语句的用法
    视图、事务
    索引
    相关子查询
    递归实现treeView下的省市联动
    创建sqlhelp类以封装对数据库的操作及对可空类型的操作
    ADO.Net操作数据库
    sql的case语句
    vue父组件异步数据子组件接收遇到的坑
    第一次用angularJS做后台管理点滴
  • 原文地址:https://www.cnblogs.com/grglym/p/7856108.html
Copyright © 2011-2022 走看看