hdu 1009 FatMouse' Trade(贪心)

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 <stdlib.h>
#include <iomanip>
using namespace std;
struct catfood
{
 int a;
 int b;
 double c;
};
int main()
{
 int M,N;
 while(cin>>M>>N)
 {
  if(M==-1&&N==-1)
   break;
  struct catfood *p,temp;
  p=(struct catfood*)malloc(sizeof(struct catfood)*N);
  int i,j,k;
  for(i=0;i<N;i++)
  {
   cin>>p[i].a>>p[i].b;
   p[i].c=1.0*p[i].a/p[i].b;
  }
  for(i=0;i<N-1;i++)
  {
   k=i;
   for(j=i+1;j<N;j++)
    if(p[j].c>p[k].c)
     k=j;
   temp=p[k];
   p[k]=p[i];
   p[i]=temp;
  }
  double sum=0;
  for(i=0;i<N;i++)
  {
   if(M>=p[i].b)
   {
    sum=sum+p[i].a;
    M=M-p[i].b;
   }
   else
   {
    sum=sum+M*1.0*p[i].a/p[i].b;
    M=0;
   }
   if(M==0)
    break;
  }
  cout<<setiosflags(ios::fixed)<<setprecision(3);
  cout<<sum<<endl;
 }
 return 0;
}