FatMouse' Trade

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

Sample Output

13.33331.500

#include<stdio.h>#include<stdlib.h>int main(){     int m,n,i,f[1002],j[1002],k,r,q;     double c[1002],sum,temp,t;     while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1)     {         for(i=0;i<n;i++)         {             scanf("%d%d",&j[i],&f[i]);             c[i]=j[i]*1.0/f[i];         }        for(i=0;i<n-1;i++)        {            for(k=0;k<n-1-i;k++)                if(c[k]<c[k+1])            {                t=c[k+1];c[k+1]=c[k];c[k]=t;                r=f[k+1];f[k+1]=f[k];f[k]=r;                q=j[k+1];j[k+1]=j[k];j[k]=q;            }        }        sum = 0;         temp = m;         for(i=0; i<n; i++)         {             temp = temp - f[i];             if(temp >= 0) sum = sum + 1.0*j[i];             else {sum = sum + (temp + f[i])*c[i]; break;}         }          printf("%.3lf\n", sum);     }}