1009 FatMouse' Trade

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<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int m,n,x,y;
    double z;
    cin>>m>>n;
    while(m!=-1&&n!=-1)
    {
        double s=0;
        int F[1000],J[10000];
        double Q[10000];
        for(int i=0;i<n;i++)
        {
        cin>>J[i]>>F[i];
        if(J[i]==0&&F[i]==0)
            Q[i]=0;
        else if(J[i]==0)
            Q[i]=0;
        else if(F[i]==0)
            Q[i]=1000000000;
        else
            Q[i]=J[i]/(double)F[i];
    }
    for(int i=0;i<n-1;i++)
        for(int j=i+1;j<n;j++)
        {
            if(Q[i]<Q[j])
            {
                z=Q[i];
                Q[i]=Q[j];
                Q[j]=z;
                x=F[i];
                F[i]=F[j];
                F[j]=x;
                y=J[i];
                J[i]=J[j];
                J[j]=y;
            }
        }//给F，J，Q全都按照Q排序 
    for(int i=0;i<n&&m!=0;i++)
    {
        if(m>F[i])
        {
            s+=J[i];
            m=m-F[i];
        }
        else 
        {
            s+=m*Q[i];
            m=0;
        }
    }//按照比例计算 
    printf("%.3f\n",s);
    cin>>m>>n;
    }
    return 0;
}