HDU_1009_FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 45970    Accepted Submission(s): 15397

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

 #include<stdio.h>
 void bubblesort(double v[],int w[],int f[],int n)
 {
     int i,j;
     double t;
     for(i=;i<=n;i++)
     {
         for(j=i+;j<=n;j++)
         {
             if(v[i]>v[j])
             {
                 t=v[i];
                 v[i]=v[j];
                 v[j]=t;

                 t=f[i];
                 f[i]=f[j];
                 f[j]=t;

                 t=w[i];
                 w[i]=w[j];
                 w[j]=t;
             }
         }
     }
 }
 int main()
 {
     int n,m,i,w[],f[];
     double v[],sum;
     while(scanf("%d %d",&m,&n)!=EOF&&(n!=-)&&(m!=-))
     {
         for(i=;i<=n;i++)
         {
         scanf("%d %d",&w[i],&f[i]);
         v[i]=f[i]*1.0/w[i];
         }
         bubblesort(v,w,f,n);

         sum=;
         for(i=;i<=n&&v[i]<=m&&m>;i++)
         {

             if(m>=f[i])
             {
                 sum+=w[i];
                 m=m-f[i];
             }
             else
             {
                 sum+=m*1.0/v[i];
                 m=;
             }
         }
         printf("%.3lf\n",sum);
     }
 }