1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45213    Accepted Submission(s): 15137

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

 

Author

CHEN, Yue

 

 

Source

ZJCPC2004

 

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
struct ln{
    double x;
    double y;
    double weight;
}a[1005];
int cmp(const void*a,const void*b)
{
    return (*(struct ln*)a).weight<(*(struct ln*)b).weight?1:-1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        memset(a,0,sizeof(a));
        if(m==-1&&n==-1)
        break;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&a[i].x,&a[i].y);
            a[i].weight=a[i].x/a[i].y;
        }
        qsort(a,n,sizeof(struct ln),cmp);
        double sum=0;

        for(int i=0;i<n;i++)
        {
            if(m>=a[i].y)
            {
                sum+=a[i].x;
                m-=a[i].y;
            }
            else
            {
                sum+=a[i].weight*m;
                m=0;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}