FatMouse' Trade

/*

problem: FatMouse' Trade

this is greedy problem.

firstly:we should calculate the average  J[i]/F[i] and sort it

secondly: according to the m and choose the most high ones

*/

#include<iostream>

#include<vector>

#include<algorithm>

using namespace std;



struct cell

{

    double j;

    double f;

    double rate;

};

bool cmp(cell x,cell y)

{

    return x.rate > y.rate;

}

int main()

{

    int m,n;

    cell temp;

    vector<cell> vec;

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

    {

        if ((m == -1)&&(n == -1))

            break;

        double count = 0.0;

        while (n--)

        {

            cin>>temp.j>>temp.f;

            if (temp.f != 0)

            {

                temp.rate = (double)temp.j/temp.f;

                vec.push_back(temp);

            }

            else

            {

                count += temp.j;

            }



        }

        sort(vec.begin(), vec.end(), cmp);

        for (int i = 0; (i < vec.size())&&(m > 0); i++)

        {

            if (m-vec[i].f >= 0)

            {

                count += vec[i].j;

                m -= vec[i].f;

            }

            else

            {

                //count += (double)((double)(m/vec[i].f))*vec[i].j;

                double temp1 = (double)m/vec[i].f;

                count += temp1*vec[i].j;

                m = 0;

            }

        }

        vec.clear();

        printf("%.3f\n",count);

    }

}

/*此题除了要满足例子以外，还要满足一些条件才能真正算ac：

0 1

1 0

1.000



1 0

0.000



5 4

10000 5

2000 2

100 0

300 0

10400.000



数据类型用double,就这样*/