FatMouse' Trade

http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1007&cid=22619

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

Source

ZJCPC2004

#include<iostream>
using namespace std;
void app(double a[],double b[],double c[],int n)
{
    int i,k,j;
    double t;
    for(i=0;i<n;i++)
    {
        k=i;
        for(j=i+1;j<n;j++)
            if(c[j]>c[k])
                k=j;
        if(i!=k)
        {
            t=a[k];
            a[k]=a[i];
            a[i]=t;
            t=b[k];
            b[k]=b[i];
            b[i]=t;
            t=c[k];
            c[k]=c[i];
            c[i]=t;
        }
    }
}
int main()
{
    int i,n,k;
    double a[1001],b[1001],c[1001],d,sx,m;
    while(scanf("%lf %d",&m,&n)==2)
    {
        if(m==-1&&n==-1)
            break;
        for(k=0;k<n;k++)
        {
            scanf("%lf",&a[k]);
            scanf("%lf",&b[k]);
            c[k]=a[k]/(1.0*b[k]);
        }
        app(a,b,c,n);
        sx=m;
        d=0;
        for(i=0;i<n;i++)
        {
            if(sx-b[i]>0)
            {
                d=d+a[i];
                sx=sx-b[i];
            }
            else
            {
                d=d+c[i]*sx;
                break;
            }
            if(sx==0)
            break;
        }
        printf("%.3f\n",d);
    }
    return 0;
}