1009

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

#include 
#include 
using namespace std;
void quickSort(int start,int eend);
int partSort(int s,int e);
double value[1000],J[1000],F[1000];
int main()
{
    int M,N;
    while(cin>>M>>N&&N<=1000&&M<=1000&&N!=-1&&M!=-1)
    {
        for(int i=0; i>J[i]>>F[i];
            value[i]=J[i]/F[i];
        }
        quickSort(0,N-1);
         double sum=0;
        for(int i=0; i0; i++)
        {
            if(M>=F[i])
            {
                sum+=J[i];
                M-=F[i];
            }
            else
            {
                sum+=M*value[i];
                M-=F[i];
            }
        }
        cout<x&&s=e)
            break;
        temp=value[s];
        value[s]=value[e];
        value[e]=temp;
        temp=J[s];
        J[s]=J[e];
        J[e]=temp;
        temp=F[s];
        F[s]=F[e];
        F[e]=temp;
    }
    value[l]=value[e];
    value[e]=x;
    temp=J[l];
    J[l]=J[e];
    J[e]=temp;
    temp=F[l];
    F[l]=F[e];
    F[e]=temp;
    return e;
}