FatMouse' Trade

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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

#include <iostream>
#include <iomanip>
#include <algorithm>

using namespace std;

struct JavaBean
{
    double F;
    double J;
};

bool cmp(JavaBean x1, JavaBean x2)
{
       return x1.J/x1.F>x2.J/x2.F;
}

JavaBean X[1010];
int main ( )
{
    int M, N, i;
    while (cin >> M >> N)
    {
        if (M == -1 && N == -1)    break;  //当输入为-1时结束程序
        for (i = 0; i < N; i++)    cin >> X[i].J >> X[i].F;//将输入放入数组中
        sort(X, X+N, cmp);
        
        double sum = 0.0;
        for (i = 0; i < N; i++)
        {
           // cout << X[i].J << " " << X[i].F << endl;
           //之前用的是if - else if 语句，一直不对，注意运行问题
            if (M > X[i].F)     { sum = sum + X[i].J;  M -= X[i].F; }
            else
            { sum = sum + X[i].J * M / X[i].F;  break; }
        }
        cout << setiosflags(ios::fixed) << setprecision(3) << sum << endl;
    }
    return 0;
}