FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 21 Accepted Submission(s) : 13

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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

Sample Output

13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int main()
 6 {
 7     int M,N,i,J[10000],F[10000],k,j,tmp,sign;
 8     double SUM;
 9     while(scanf("%d%d",&M,&N)!=EOF)
10     {
11         if(M==-1&&N==-1)
12             break;
13         k=0;SUM=0;
14         while(N--)
15         {
16             scanf("%d%d",&J[k],&F[k]);
17             k++;
18         }
19         for(i=k-1;i>=0;i--)
20         {
21              sign=i;
22              for(j=i-1;j>=0;j--)
23                 if(J[sign]*F[j]>J[j]*F[sign])sign=j;
24              if(sign!=i)
25              {
26                  tmp=J[i];
27                  J[i]=J[sign];
28                  J[sign]=tmp;
29 
30                 tmp=F[i];
31                  F[i]=F[sign];
32                  F[sign]=tmp;
33              }
34         }
35 
36         for(i=0;i<k;i++)
37         {
38             M-=F[i];
39             if(M<0)
40                 {SUM+=((J[i]*1.0)/F[i])*(M+F[i]);break;}
41             SUM+=J[i];
42         }
43         printf("%.3lf
",SUM);
44 
45     }
46     return 0;
47 }

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