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
#include<iostream>
#include<iomanip>
using namespace std;
struct aa
{int a,b;
double c;
}ab[1002]; 
int main()
{
	int m,n,i,k,l,x,a1,b1,max=1001;
	double a,b,c,y,c1;
	while(cin>>m>>n&&m!=-1&&n!=-1)
	{
		k=0;
		for(i=0;i<n;i++)
		{cin>>a>>b;
		if (b == 0)  
                                c = static_cast<double>(max);//////这句不明白  
		else
		c=a/b;
		 if(k==0)
		 {ab[0].a=a;ab[0].b=b;ab[0].c=c;k=1;}
		 else 
		 {
			 for(l=0;l<i;l++)
				 if(ab[l].c>c) break;
				 
				 for(x=i;x>l;x--)
				 {ab[x].a=ab[x-1].a;
				 ab[x].b=ab[x-1].b;
				 ab[x].c=ab[x-1].c;
				 }
				 ab[l].a=a;
				 ab[l].b=b;
				 ab[l].c=c;
		 }
		}
y=0;
       while(m>0&&n>0)
	   {
		   n--;
		   if(ab[n].b<m)
			   y=y+ab[n].a;
		   else
		   { y=y+ab[n].c*m;break;}
		   m=m-ab[n].b;
	   }
	  cout<<setiosflags(ios::fixed)<<setprecision(3)<<y<<endl;

	}

return 0;
}