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[1005];

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==-1&&n==-1)
            break;
        for(int i = 0; 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 = 0;

        for(int i = 0; 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 0;
}