FatMouse' Trade

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

 1 #include <stdio.h> 
 2 #define N 1001
 3 
 4 int main(){
 5     int m;
 6     int n;
 7     int i;
 8     int j;
 9     int J[N];
10     int F[N];
11     double unit[N];
12     double temp;
13     double get;
14 
15     while(1){
16         scanf("%d%d",&m,&n);
17 
18         if(m==-1 && n==-1)
19             break;
20 
21         get=0;
22         for(i=0;i<n;i++){
23             scanf("%d%d",&J[i],&F[i]);
24             unit[i]=(double)J[i]/F[i];
25         }
26 
27         for(i=0;i<n-1;i++){
28             for(j=i+1;j<n;j++){
29                 if(unit[i]<unit[j]){
30                     temp=unit[i];
31                     unit[i]=unit[j];
32                     unit[j]=temp;
33 
34                     temp=J[i];
35                     J[i]=J[j];
36                     J[j]=temp;
37 
38                     temp=F[i];
39                     F[i]=F[j];
40                     F[j]=temp;
41                 }
42             }
43         }
44 
45         
46         for(i=0;i<n;i++){
47             if(m>=F[i]){
48                 m=m-F[i];
49                 get+=J[i];   //这里J[i]改成unit[i]*F[i]会丢失精度导致出错
50 
51                 if(m==0)
52                     break;
53             }
54 
55             else{
56                 get+=unit[i]*m;
57                     
58                 break;
59             }
60         }
61         printf("%.3lf
",get);
62     }    
63     return 0;
64 }