hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12798    Accepted Submission(s): 3758

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
  
  

 

 

Author

CHEN, Yue

 

 

Source

ZJCPC2004

 

 

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#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
    int j,f;
    double av;
};
node a[1010];
bool cmp(node x,node y)
{
    return x.av>y.av;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&m,&n)==2)
    {
        if(n==-1&&m==-1) break;
        for(int i=0;i<n;i++) {scanf("%d%d",&a[i].j,&a[i].f);a[i].av=1.0*a[i].j/a[i].f;}
        sort(a,a+n,cmp);
        double cnt=0;
        double ans=0;
        for(int i=0;i<n;i++)
        {
           //cout<<a[i].j<<".."<<a[i].f<<"..."<<a[i].av<<endl;
            if(a[i].f<=m-cnt) cnt+=a[i].f,ans+=a[i].j;
            else
            {
                ans+=(m-cnt)/a[i].f*a[i].j;
                break;
            }
        }
        printf("%.3lf/n",ans);
    }
    return 0;
}