hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62744    Accepted Submission(s): 21181

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

很简单的贪心，直接上代码了；

#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;
struct room{
	double JavaBeans;
	double catfoods;
	double rate;
};
bool compare(room a,room b)
{
	return a.rate>b.rate;
}
int main()
{
	int m,n;
	int i;
	room room[1001];
	while(cin>>m>>n)
	{
		if(m==(-1)&&n==(-1))
		{
			return 0;
		}
		double s=0;
		for(i=0;i<n;i++)
		{
			cin>>room[i].JavaBeans>>room[i].catfoods;
			room[i].rate=(room[i].JavaBeans/room[i].catfoods);
		}
		sort(room,room+n,compare);
		for(int k=0;k<n;k++)
		{
			if(m>room[k].catfoods)
			{
				s=s+room[k].JavaBeans;
				m=m-room[k].catfoods;
			}
			else
			{
				s=s+(double)m/room[k].catfoods*room[k].JavaBeans;
				break;
			}
		}
		cout.setf(ios::fixed);
		cout << setprecision(3) << s << endl;
	}
	return 0;
}