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=1;i<=n;i++)
    {
        for(j=i+1;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[1000],f[1000];
    double v[1000],sum;
    while(scanf("%d %d",&m,&n)!=EOF&&(n!=-1)&&(m!=-1))
    {
        for(i=1;i<=n;i++)
        {
        scanf("%d %d",&w[i],&f[i]);
        v[i]=f[i]*1.0/w[i];
        }
        bubblesort(v,w,f,n);
        
        sum=0;
        for(i=1;i<=n&&v[i]<=m&&m>0;i++) 
        {
             
            if(m>=f[i]) 
            {
                sum+=w[i];
                m=m-f[i];
            }
            else
            {
                sum+=m*1.0/v[i]; 
                m=0;
            }  
        }
        printf("%.3lf
",sum);     
    }
}