hdu_1009_FatMouse' Trade_201310280910

FatMouse' Trade

http://acm.hdu.edu.cn/showproblem.php?pid=1009

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 35250    Accepted Submission(s): 11553

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>

 typedef struct ST
 {
     int j;
     int f;
     double t;
 }ST;
 ST s[];

 int cmp(const void *a,const void *b)
 {
     return (*(ST *)a).t > (*(ST *)b).t ?  : -;
 }

 int main()
 {
     int m,n;
     while(scanf("%d %d",&m,&n),(m!=-&&n!=-))
     {
         int i,j;
         int num;
         double sum=;
         for(i=;i<n;i++)
         {
             scanf("%d %d",&s[i].j,&s[i].f);
             s[i].t = s[i].j*1.0/s[i].f;
         }
         qsort(s,n,sizeof(s[]),cmp);
         num=m;
         for(i=n-;i>=;i--)
         {
             if(num>s[i].f)
             {
                 sum+=s[i].j;
                 num-=s[i].f;
             }
             else
             {
                 sum+=s[i].t * num;
                 num=;
             }
             if(num==)
             break;
         }
         printf("%.3lf\n",sum);
     }
     return ;
 }