求解一道ACM的题

题目：
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

我的程序：

#include<iostream>

using namespace std;



typedef struct node

{

	int f,j;

	float info;

}node;



int main()

{

	int m,n,i,j;

	node st[2000],temp;

	float result;

	while(cin>>m>>n&&m!=-1&&n!=-1)

	{

		result=0;

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

		{

			cin>>st[i].f>>st[i].j;

			if(st[i].j==0)

				st[i].info=1000000000;

			else

				st[i].info=float(st[i].f)/float(st[i].j);

		}

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

			for(j=0;j<n-1-i;j++)

			{

				if(st[j].info<st[j+1].info)

				{

					temp=st[j];

					st[j]=st[j+1];

					st[j+1]=temp;

				}

			}

		i=0;

		while(i<n)

		{

			if(m>=st[i].j)

			{

				result=result+st[i].f;

				m=m-st[i].j;

				i++;

			}

			else

			{

				result=result+st[i].info*m;

				break;

			}

		}

		printf("%0.3f\n",result);

	}

	return 0;

}

通不过测试，显示是：wrong answer
求助错哪里了，或者哪里考虑不周全了