hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73443    Accepted Submission(s): 25220

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<math.h>
#include<string>
#include<algorithm>
using namespace std;
struct node
{
	int x,y;
	double avg;
}a[1001];
bool cmp(node a,node b)
{
	return a.avg>b.avg;
}
int main()
{ 
     int m,n;
     while(cin>>m>>n)
     {
     	double sum=0;
     	if(m==-1&&n==-1) break;
     	for(int i=0;i<n;i++)
     	{
     		cin>>a[i].x>>a[i].y;
     		a[i].avg=(a[i].x*1.0/a[i].y);
		}
		sort(a,a+n,cmp);
		int i=0;
		while(i<n&&m!=0)
		{
			if(a[i].y<=m) 
			{	sum=a[i].x+sum;
				m-=a[i].y;
			//	cout<<sum<<endl;
			}
			else 
			{ 
			
			 	sum+=m*a[i].avg;
			 	m=0;
			}	
			i++; 	
		 
		}
		printf("%.3lf\n",sum);					
	}
}