FatMouse' Trade（hdoj1009）（简单贪心）

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

#include<stdio.h>
#include<algorithm>
using namespace std;

struct st
{
    double l;
    double v;
}s[1002];

int cmp(st a,st b)
{
    return a.v/a.l>b.v/b.l;
} 
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf %lf",&s[i].v,&s[i].l);
        }
        sort(s+1,s+n+1,cmp);
        double ans=0,sum=0;
        for(int i=1;i<=n;i++)
        {
            if(ans+s[i].l<=m)
            {
                ans+=s[i].l;
                sum+=s[i].v;
            }
            else
            {
                sum=sum+(m-ans)*s[i].v/s[i].l;
                break;
            }
        } 
        printf("%.3lf\n",sum);
    }
}