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.

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.

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.

``````5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1``````

``````13.333
31.500``````

``````#include"stdio.h"
double a={0};
int found(int n)
{
double max=0;
int j=0,i;
for(i=0;i<n;i++)
{
if(a[i]>max)
{
max=a[i];
j=i;
}
}
if(max==0)
return -1;
else
return j;

}
int main()
{
double m,sum=0;
int n,t=0,i;
int f, j;
scanf("%lf %d",&m,&n);
while(m!=-1&&n!=-1)
{
for(i=0;i<n;i++)
{
scanf("%lf %lf",&a[i],&a[i]);
a[i]=a[i]/a[i];
}
while(m!=0)
{
t=found(n);
if(t==-1)
{
break;
}
if(m>=a[t])
{
m-=a[t];
sum+=a[t];
a[t]=0;
}
else
{
sum+=((double)m)/((double)a[t])*a[t];
break;
}
}
printf("%.3lf\n",sum);
scanf("%lf %d",&m,&n);
sum=0;

}

return 0;
}
``````

``````#include"stdio.h"
double a={0};
int found(int n)
{
double max=0;
int j=0,i;
for(i=0;i<n;i++)
{
if(a[i]>max)
{
max=a[i];
j=i;
}
}
if(max==0)
return -1;
else
return j;

}
int main()
{
double m,sum=0;
int n,t=0,i;
int f, j;
scanf("%lf %d",&m,&n);
while(m!=-1&&n!=-1)
{
for(i=0;i<n;i++)
{
scanf("%lf %lf",&a[i],&a[i]);
a[i]=a[i]/a[i];
}
while(m!=0)
{
t=found(n);
if(t==-1)
{
break;
}
if(m>=a[t])
{
m-=a[t];
sum+=a[t];
a[t]=0;
}
else
{
sum+=((double)m)/((double)a[t])*a[t];
break;
}
}
printf("%.3lf\n",sum);
scanf("%lf %d",&m,&n);
sum=0;

}

return 0;
}
``````