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 ii pounds of JavaBeans and requires F ii pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J ii* a% pounds of JavaBeans if he pays F ii* 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 ii and F ii 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

贪心，一直贪性价比最高的，很奇怪我用cin一直是wa，怀疑cin和scanf读入的精度不一样

#include
#include
#include
#include
using namespace std;
struct node
{
    double x,y;
    double v;
};
node a[1005];
bool cmp(node a,node b)
{
    return a.v>b.v;
}
int main()
{
    double sum,ans;
    int n,k,i;
    while(scanf("%lf%d",&sum,&n))
    {
        if (sum==-1&&n==-1) break;
        ans=0;
        k=0;
        for(i=0;i>a[i].x>>a[i].y;
            a[i].v=a[i].x/a[i].y;
        }
        sort(a,a+n,cmp);
        while(sum>0&&ka[k].y)
            {
                sum=sum-a[k].y;
                ans=ans+a[k].x;
                k++;
            }
            else
            {
                ans=ans+sum/a[k].y*a[k].x;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}