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[];
 int cmp(const void*a,const void*b)
 {
     return (*(struct ln*)a).weight<(*(struct ln*)b).weight?:-;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     int m,n;
     while(~scanf("%d%d",&m,&n))
     {
         memset(a,,sizeof(a));
         if(m==-&&n==-)
         break;
         for(int i=;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=;

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