hdoj1009FatMouse' Trade(贪心求解)

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<stdio.h>
#include<stdlib.h>
double J[1005],F[1005];
struct node
{
    double s;
    int k;
}S[1000];
int cmp(const void* p1,const void* p2)
{
    return ((node *)p2)->s > ((node *)p1)->s ? 1 : -1;//之前出错,因为这里写成((node*)p2)->s-((node*)p1)->s;
}
int main()
{
    double m;
    int n;
    while(scanf("%lf%d",&m,&n)!=EOF)
    {
        if(m==-1&&n==-1) break;
        for(int i=0;i<n;i++)
        { scanf("%lf%lf",&J[i],&F[i]);
          S[i].s=J[i]/F[i];
          S[i].k=i;
        }
        qsort(S,n,sizeof(S[0]),cmp);
        double f=0;
        int j=0;
        for(int j=0;j<n;j++)
        {
           if(F[S[j].k]<=m)
           {
               f=f+J[S[j].k];
               m=m-F[S[j].k];
           }
           else
           {
               double k1=m/F[S[j].k];
               f=f+J[S[j].k]*k1;
               break;
           }
        }
        printf("%.3lf/n",f);
    }
    return 0;
}