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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output
13.333
31.500

import java.text.DecimalFormat;
import java.util.*;

public class FatMouseTrade {
    public static void main(String args[]) {
        Scanner input = new Scanner(System.in);
        boolean flag = true;
        while(flag){
            int a = input.nextInt();
            int b = input.nextInt();
            if (a != -1 && b != -1) {
                int[][] array = new int[b][2];
                for (int i = 0; i < b; i++) {
                    array[i][0] = input.nextInt();
                    array[i][1] = input.nextInt();
                }
                System.out.println(getMaxBean(array,a));
            }else{
                flag  =false;
            }
        }
    }

    public static String getMaxBean(int[][] a, int b) {
        float maxBean = 0.000f;
        DecimalFormat   df   =   new   DecimalFormat("##0.000");
        for (int i = 0; i < a.length; i++) {
            for (int j = 1; j < a.length - i; j++) {
                if (((float)a[j - 1][0] / (float)a[j - 1][1]) < ((float)a[j][0] / (float)a[j][1])) {
                    int tempi = a[j - 1][0];
                    a[j - 1][0] = a[j][0];
                    a[j][0] = tempi;
                    int tempj = a[j - 1][1];
                    a[j - 1][1] = a[j][1];
                    a[j][1] = tempj;
                }
            }
        }

        int flag=0;
        while(b >= 0 && flag<a.length){
            if(b >= a[flag][1]){
                maxBean += a[flag][0];
                b -= a[flag][1];
                flag++;
             //   System.out.println("---------"+maxBean);
            }else{
                maxBean =maxBean + ((float)a[flag][0] / (float)a[flag][1])*b;
                b -= a[flag][1];
                flag++;
                //System.out.println("*********"+maxBean);
            }
        }
        return df.format(maxBean);

    }


}

并没有ac，但是能想到的测试数据都可以通过
测试数据：
0 1
1 0
0.000

0 0
0.000

0 3
2 0
3 1
4 0
6.000

2 3
1 5
1 4
1 6
0.500

0 3
1 3
2 3
3 3
0.000

0 1
2 0
2.000

5 4
10000 5
2000 2
100 0
300 0
10400.000

2 2
3 1
2 3
3.667