FatMouse' Trade

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

题解：我建立了一个结构体，然后将食物价值用r表示，然后把结构体排序，依次从最大的开始递减，可以保证价值最大。

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<vector>

using namespace std;

struct food

{

    double j;

    double f;

    double r;

}fo[];

bool compare(food a,food b)

{

    return a.r>b.r;

}

int main()

{

    int m,n;

    while(scanf("%d %d",&m,&n),m!=-,n!=-)

    {

        for(int i=;i<n;i++)

        {

            scanf("%lf %lf",&fo[i].j,&fo[i].f);

            fo[i].r=fo[i].j/fo[i].f;

        }

        sort(fo,fo+n,compare);

        double ans=;

        for(int i=;i<n;i++)

        {

            if(m>=fo[i].f)

            {

                ans+=fo[i].j;

                m-=fo[i].f;

            }

            else

            {

                ans+=m*fo[i].r;

                break;

            }

        }

        printf("%.3lf\n",ans);

    }

    return ;

}