FatMouse' Trade

 FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 
     

 

#include<iostream>
using namespace std;
struct room
{
	double catfood;
	double mousefood;
	double aver;
};
int main()
{
  int M,N,i,j,k;
  while(cin>>M>>N)
  {
	  room a[1000];
	  double sum=0;
	  if(M==-1&&N==-1)
		  break;
	  else
	  {   
		  for(i=0;i<N;i++)
		  {
			  cin>>a[i].mousefood>>a[i].catfood;
		  }
		  for(i=0;i<N;i++)
		  {
			  a[i].aver=a[i].mousefood/a[i].catfood;//相同猫食(消耗)时候可以得到的鼠食(利益)
		  }
         for(i=0;i<N-1;i++)
		 {
			 j=i;
			 for(k=j+1;k<N;k++)
			 {
				 if(a[k].aver>a[j].aver)
				 {
					 j=k;
				 }
			 }
			 if(i!=j)
			 {
			 room t;   //相同消耗，得到的最多 的降序排列
			 t=a[j];
			 a[j]=a[i];
			 a[i]=t;
			 }
		 }
		 for(i=0;i<N;i++)
		 {
			 if(a[i].catfood<M)
			 {
                 sum=sum+a[i].mousefood;
				 M=M-a[i].catfood;
			 }
              else
			  {
				  sum=sum+M*(a[i].mousefood)/a[i].catfood;
				  break;
			  }
		 }
		 printf("%.3lf\n",sum);	 
	  }

  }
  return 0;
}