1009 FatMouse' Trade（初入贪心）

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<iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <math.h>
using namespace std;
struct sa
{
    int jab;
    int food;
    double awp;
}data[2000];
int cmp(const sa &a,const sa &b)
{
    return (a.awp)>(b.awp);
}
int main()
{
    int m,n;
    double ans;
    while(cin>>m>>n&&m!=-1&&n!=-1)
    {
        for(int i=0;i<n;i++)
        {
            cin>>data[i].jab>>data[i].food;
            data[i].awp=(double)data[i].jab/(double)data[i].food;
        }
        sort(data,data+n,cmp);
        ans=0.0;
        for(int i=0;i<n;i++)
        {
            if(m>=data[i].food)
            {
                ans+=data[i].jab;
                m=m-data[i].food;
            }
            else
            {
                ans+=m*data[i].awp;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}