FatMouse' Trade

Time Limit : 2000/1000ms(Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 6   AcceptedSubmission(s) : 4

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Problem Description

FatMouseprepared M pounds of cat food, ready to tradewith the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th roomcontains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to（不用） trade for all the JavaBeans in the room, instead, hemay get J[i]* a% pounds of JavaBeans if he paysF[i]* a% pounds of cat food. Here a is a realnumber. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The inputconsists of multiple test cases. Each test case begins with a line containingtwo non-negative integers M and N. Then N lines follow, each contains two non-negativeintegers J[i] and F[i] respectively. The last test case is followed by two -1's. All integersare not greater than 1000.

Output

For each testcase, print in a single line a real number accurateup to 3（精确到小数点后三位） decimal places, which is the maximum amount of JavaBeansthat 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 

//*************************
//* 任务介绍：交易        *
//* 作者： 何香           *
//* 完成时间：2013.10.18  *
//*************************

#include<iostream>
#include<iomanip>
using namespace std;
int main ()
{
	int M,N,i,j;//M是所有的catfood，N是一共有N个room
	double A[1002];//记录第i个room平均1 food换的javabean的数目
	int J[1002],F[1002];//Javabean的数目，catfood的数目
	while(cin>>M>>N)
	{
		if(M!=(-1)||N!=(-1))
		{
		int t1,t2;
		double t;
		//输入每个房里Javabean和food的数目
		for(i=1;i<=N;++i)
		{
			cin>>J[i]>>F[i];
			A[i]=1.0*J[i]/F[i];
		}
		//用冒泡排序法把平均值的大小由大到小排序，同时也使J和F也重新排序
		for(i=1;i<N;++i)
		{
			for(j=1;j<=N-i;++j)
				if(A[j]<A[j+1])
				{
					t=A[j],t1=J[j];t2=F[j];
					A[j]=A[j+1],J[j]=J[j+1],F[j]=F[j+1];
					A[j+1]=t,J[j+1]=t1,F[j+1]=t2;
				}
		}
		
		int m=0;//记录所加的catfood的数目
		int temp;
		double max_J=0;//记录最大的Javabean数目
		
		//计算达到M时最大的Javabean
/*		for(i=1;i<=N;++i)
		{
			m=temp;
			m=m+F[i];
			if(m<=M)
			{
				max_J=max_J+J[i];
			}
				if(m=M)
				{
					max_J=max_J+J[i];
					break;
				}
			else
			{
				t=1.0*(M-temp)/F[i];
				max_J=max_J+J[i]*t;
				break;
			}
		}*/

		for(i=1;i<=N;++i)
		{
			if(M>=F[i])
			{
				max_J+=J[i];
				M-=F[i];
			}
			else
			{
				t=1.0*M/F[i];
				max_J+=J[i]*t;
				M=0;

			}
		}
		cout<<setiosflags(ios::fixed)<<setprecision(3);
		cout<<max_J<<endl;
		}
		else
			break;
	}
	
	return 0;
}