hdu 1009 FatMouse' Trade

Posted 2020-07-08 行尸走肉

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

ZJCPC2004

 

#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\\n",ans);
    }
}

View Code

简单的贪心，没什么好说的。