• hdu 1009 FatMouse' Trade


    FatMouse' Trade

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 63732    Accepted Submission(s): 21565


    Problem Description
    FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
    The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.
     
    Input
    The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.
     
    Output
    For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.
     
    Sample Input
    5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
     
    Sample Output
    13.333 31.500
     
    Author
    CHEN, Yue
     
    Source
     
    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    using namespace std;
    struct Node
    {
        double f,j;
        double wei;
    }room[100005];
    bool cmp(Node a,Node b)
    {
        return a.wei>b.wei;
    }
    int main()
    {
        double m;
        int n;
        while(scanf("%lf%d",&m,&n)&&(m!=-1&&n!=-1))
        {
            for(int i=0;i<n;i++)
            {
                scanf("%lf%lf",&room[i].j,&room[i].f);
                room[i].wei=room[i].j/room[i].f;
            }
            sort(room,room+n,cmp);
            double ans=0;
            for(int i=0;i<n;i++)
            {
                if(m<=0) break;
                if(m>=room[i].f) {m-=room[i].f;ans+=room[i].j;}
                else if(m<room[i].f){ans+=m*room[i].wei;m-=m;break;}
            }
            printf("%.3lf
    ",ans);
        }
    }
    View Code

    简单的贪心,没什么好说的。

  • 相关阅读:
    计算机与软件工程 作业一
    软件工程作业6 咸鱼不闲团队 校园二手交易系统
    软件工程 作业5
    计算与软件工程 作业4
    软件工程第四次作业评价总结
    计算与软件工程 作业3
    计算与软件共程作业2
    计算与软件工程 作业一
    计算与软件工程作业六
    计算与软件工程作业五
  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5506732.html
Copyright © 2020-2023  润新知