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

 

Sample Output

#include<iostream>

#include<string>

#include<algorithm>

#include<cmath>

#include<cstdio>

using namespace std;

struct food

{

	double ji;

	double fi;

	double xingjiabi;//按照性价比排序

};

bool cmp(const food &a,const food &b)

{

	if(a.xingjiabi!=b.xingjiabi)

		return a.xingjiabi>b.xingjiabi;//性价比高的排前面

	else

		return a.fi<b.fi;//性价比一样则需要代价的fi（猫粮）小的排前面，不过再看看我后面的代码似乎不需要这条

}

int main(void)

{

	int m,n;

	double ans,rest;

	while (cin>>m>>n)

	{

		if(m==-1&&n==-1)

			break;

		food *list=new food[n];

		for (int i=0; i<n; i++)

		{

			scanf("%lf %lf",&list[i].ji,&list[i].fi);

			list[i].xingjiabi=list[i].ji*1.0/list[i].fi;

		}

		sort(list,list+n,cmp);

		ans=0;

		rest=m;

		for (int i=0; i<n; i++)

		{

			if(rest>=list[i].fi)//可以全部拿到直接相加

			{

				rest-=list[i].fi;

				ans+=list[i].ji;

			}

			else

			{

				ans=ans+list[i].xingjiabi*rest;//只能拿到一部分则把剩下的部分全给了，按照付出的百分比进行加成

				rest-=rest;

			}

		}

		printf("%.3lf\n",ans);

		delete []list;

	}

	return 0;

}