ACM FatMouse' Trade

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. 

InputThe 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. 
OutputFor 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

 #include<bits/stdc++.h>
 using namespace std;
 struct node{
     int j;  /*理想下越大越好*/
     int f;   /*理想下越小越好*/
     double rate;  /*性价比*/
 }jb[];
 
 bool cmp(node a,node b)
 {
     if(a.rate != b.rate)
         return a.rate > b.rate;
     else
         return a.f < b.f;
 }
 
 int main()
 {
     int m,n;
     while(cin>>m>>n)  /*m(固定的猫粮) n（房间数）*/
     {
         if(m==-&&n==-)
             break;
         for(int i = ; i< n; i++)
         {
               scanf("%d%d",&jb[i].j,&jb[i].f); /*j（房间内最多的JB数量）f(换购猫粮)*/
               jb[i].rate = jb[i].j*1.0/jb[i].f; /*记得在分子*1.0，否则出来的答案只能是int类型的*/
         }
         sort(jb,jb+n,cmp);
         double ans = ;
 
         for(int i = ; i < n; i++)
         {
             //cout<<jb[i].rate<<" "<<jb[i].f<<endl;
             if(m >= jb[i].f)
             {
                 ans+=jb[i].j;
                 m -= jb[i].f;
             }else{
                 ans += jb[i].rate*m;   /*根据计算公式推导*/
                 break;
             }
         }
         printf("%.3lf\n",ans);
     }
     return ;
 }