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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500 题目意思：大老鼠一共有m磅的猫粮要和n个房间里面的猫做交易来换取他喜欢吃的食物，他不需要和所有的房间都进行交易，他支出F[i]，会得到J[i], 问他能得到的最大食物数。 解题思路：这算是我接触到的第一道贪心策略的题目，我用结构体来记录每个房间需要交易的猫粮和能换到的食物，以及兑换比率，贪心策略就是每次选择兑换比率大的来兑换 就能够得到最多兑换食物量。 上代码：

#include
     
      #include
      
       using namespace std;struct food{    int j;    int f;    double q;};int compare(food a,food b){    if(a.q>b.q)        return 1;    else        return 0;}int main(){    int n,m,i,k,l;    double s;    struct food a[1000],t;    while(scanf("%d%d",&m,&n)!=EOF)    {        if(m==-1||n==-1)            break;        for(i=0; i
       
        =a[i].f)            {                s=s+a[i].j;                m=m-a[i].f;            }            else if(m
        
         =0)            {                s=s+(m*a[i].q);                break;            }        }        printf("%.3lf\n",s);    }    return 0;}