HDU1009

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
   //  先找出每个的汇率a.j/a.f，然后从小到大排出，如果列出的a.f比m大，就能直接算出，比他小就乘上汇率#include <iostream>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef struct{    int j,f;    double r;} node;node a[10000],temp;int main(){    int m,n,i;    while(scanf("%d%d",&m,&n)!=EOF&&(n!=-1||m!=-1))    {        for(i=0; i<n; i++)        {            scanf("%d%d",&a[i].j,&a[i].f);            a[i].r=1.0*a[i].j/a[i].f;        }        int e;        for(i=0; i<n-1; i++)            for(e=i+1; e<n; e++)                if(a[i].r<a[e].r)                {                    temp=a[i];                    a[i]=a[e];                    a[e]=temp;                }        double sum=0;        for(i=0; i<n; i++)            if(m>=a[i].f)            {                sum+=a[i].j;                m-=a[i].f;            }            else            {                sum+=m*a[i].r;                m=0;                break;            }        printf("%.3lf\n",sum);    }    return 0;}