hdu1009

 

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23490    Accepted Submission(s): 7380

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<stdio.h>
#include<algorithm>
using namespace std;
struct Trade
{
    int j,f;
    double percent;
}mouse[3001];
bool cmp(Trade a,Trade b)
{
    return a.percent>b.percent;
}

int main()
{
    int n,m;
    while(scanf("%d%d",&m,&n)!=EOF&&(n!=-1||m!=-1)){
        int i;
        for(i=0;i<n;i++){
            scanf("%d%d",&mouse[i].j,&mouse[i].f);
            mouse[i].percent=(double)mouse[i].j/mouse[i].f;
        }
        sort(mouse,mouse+n,cmp);
        double sum=0;
        for(i=0;i<n;i++)
        {
            if(m>mouse[i].f){
                     sum+=mouse[i].j;
                     m-=mouse[i].f;
                 }
            else{
                sum+=mouse[i].percent*m;
                m=0;
                break;
            }
                    
                       
        }
       
        printf("%.3lf\n",sum);
       
    }
    return 0;
}