HDOJ 1009 FatMouse' TradeD

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
   
   
   
   
   
   
   
   

    
    
    
    
//
//  main.cpp
//  FatMouse' Trade
//
//  Created by 张嘉韬 on 16/3/8.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//

#include <iostream>
#include <cstring>
#include <algorithm>
const int maxn=1010;
using namespace std;
struct room
{
    double j;
    double f;
    double a;
}rooms[maxn];
int cmp(const void *a,const void *b)
{
    return (*(room *)a).a>(*(room *)b).a?-1:1;
}
int main(int argc, const char * argv[]) {
    int m,n;
    while(scanf("%d%d",&m,&n)==2)
    {
        double sum=0;
        if(m==-1&&n==-1) break;
        else
        {
            for(int i=0;i<n;i++)
            {
                cin>>rooms[i].j>>rooms[i].f;
                rooms[i].a=rooms[i].j/rooms[i].f;
            }
            qsort(rooms,n,sizeof(rooms[0]),cmp);
//            for(int i=0;i<n;i++) cout<<rooms[i].a<<endl;
            for(int i=0;i<n;i++)
            {
               if(m>=rooms[i].f)
               {
                   sum+=rooms[i].j;
                   m=m-rooms[i].f;
               }
                else
                {
                    sum+=rooms[i].a*m;
                    //cout<<"("<<sum<<")"<<endl;
                    break;
                }
                //cout<<"("<<sum<<")"<<endl;
                
            }
            //cout<<sum<<endl;
            printf("%.3f",sum);
            cout<<endl;
        }
    }
    return 0;
}