1009：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

 

 

代碼實現：

#include<cstdio>

double java[100000],f[100000],a[100000];

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