HDU 1009 fatmouse' trade(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

code

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

struct goods
{
    double j;
    double f;
    double s;

};

bool cmp(const goods &A,const goods &B)
{
    return A.s>B.s;
}
int main()
{
    goods buf[1000];
    double m;
    int n;
    while(cin>>m>>n)
    {
        if(m==-1&&n==-1)break;
        for(int i=0;i<n;i++)
        {
            cin>>buf[i].j>>buf[i].f;
            buf[i].s=buf[i].j/buf[i].f;
        }
        sort(buf,buf+n,cmp);
        int id=0;
        double ans=0;
        while(m>0&&id<n)
        {
            if(m>buf[id].f)
            {
                ans+=buf[id].j;
                m-=buf[id].f;
            }
            else
            {
                ans+=buf[id].j*m/buf[id].f;
                m=0;
            }
            id++;
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}

思路：
求出每个房间一磅猫粮能换多少食品，从高到低排序，先换最划算的。