杭电 acm 1009 FatMouse' Trade

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48150    Accepted Submission(s): 16204

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

 

//一开始看不懂题意，其实就是有n个猫住的小房子；n列数据即可理解为价位表，
//如几个猫食可以兑换几个豆子，用老鼠持有的猫食总数m从兑换率最高的房间开始依次兑换：贪心思想；
#include
#include
using namespace std;

struct food
{
	int beans;
	int catfood;
	double rate;
};

bool cmp(const food a,const food b)//对结构中某一数据进行排序；
{
	return a.rate>b.rate;
}

int main()
{
	int m,n,i;
	double s;

	while(cin>>m>>n)
	{	
		if(m==-1&&n==-1)
			break;
		food *p=new food[n];
		s=0;
		for(i=0;i>p[i].beans>>p[i].catfood;
			p[i].rate=(double)p[i].beans/p[i].catfood;
		}
	
	sort(p,p+n,cmp);
	for(i=0;i