Who's in the Middle

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<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<deque>#include<list>using namespace std;struct Node{    double j,f,p;} node[10000];int cmp(Node x,Node y){    return x.p>y.p;}int main(){    int m,n;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==-1 || m==-1)        break;        else        {            double sum = 0,max = 0;            int i,j;            for(i = 0; i<m; i++)            {                scanf("%lf%lf",&node[i].j,&node[i].f);                node[i].p = node[i].j/node[i].f;            }            sort(node,node+m,cmp);            for(i = 0; i<m; i++)            {                if(n>node[i].f)                {                    sum+=node[i].j;                    n-=node[i].f;                }                else                {                    sum+=node[i].p*n;                    break;                }            }            printf("%.3lf\n",sum);        }    }    return 0;}