FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37292    Accepted Submission(s): 12311

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<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
struct info
{
    int j;
    int f;
    double ave;
}unit[10005];
int cmd(const info &x,const info &y)
{
    //printf("fhdshfjks");
    return x.ave<y.ave;
}
int main()
{
    int i,j,m,n;
    double sum;
    //info unit[10005];
    while(scanf("%d%d",&m,&n)!=EOF)
    {
     if((n==-1)&&(m==-1))
     break;
     for(i=0;i<n;i++)
     {
         scanf("%d%d",&unit[i].j,&unit[i].f);
         unit[i].ave=unit[i].f/(unit[i].j*1.0);
     }
     sort(unit,unit+n,cmd);
     //for(i=0;i<n;i++)
     //printf("%f\n",unit[i].ave);
     sum=0;
     //printf("%d",m);
     for(i=0;i<n&&m!=0;i++)
     {
         if(m>=unit[i].f)
         {
             sum=sum+unit[i].j;
             //printf("hdjks%f\n",sum);
             m=m-unit[i].f;
         }
         else
         {
             sum=sum+(m/(unit[i].f*1.0))*unit[i].j;
             //printf("aaa%f\n",sum);
             m=0;
             break;
         }

     }

     printf("%.3f\n",sum);
    }
    return 0;

}

 

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