问题描述如下：

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

算法分析：

该算法采用先对兑换比例进行排序；然后再从比率最高的进行兑换；直到兑换完为止。

参考代码：

public static void main(String[] args) {
		// TODO Auto-generated method stub
		int m, n, i, j, tempa, tempb;
		float temp;
		float sum = 0;

		Scanner scanner = new Scanner(System.in);
		System.out.println("请输入有多少猫粮：");
		m = scanner.nextInt();

		System.out.println("请输入有多少房间：");
		n = scanner.nextInt();

		System.out.println("猫粮：" + m + "\t房间：" + n);
		int a[] = new int[n];
		int b[] = new int[n];
		float r[] = new float[n];
		System.out.println("依次输入可兑换的javabean，和需要的猫粮：");

		for (i = 0; i < n; i++) {
			a[i] = scanner.nextInt();
			b[i] = scanner.nextInt();
			r[i] = (float) a[i] / b[i];
		}

		for (i = 0; i < n; i++) {
			System.out.print(a[i] + "---");
			System.out.println(b[i]);
		}
		for (i = 0; i < n - 1; i++) {
			for (j = 0; j < n - i - 1; j++)
				if (r[j] < r[j + 1]) {
					temp = r[j];
					r[j] = r[j + 1];
					r[j + 1] = temp;
					tempa = a[j];
					a[j] = a[j + 1];
					a[j + 1] = tempa;
					tempb = b[j];
					b[j] = b[j + 1];
					b[j + 1] = tempb;

				}
		}
		for (i = 0; i < n; i++) {
			System.out.print(r[i] + "---");
			System.out.print(a[i] + "---");
			System.out.println(b[i]);
		}

		if (m > 0) {
			for (i = 0; i < b.length; i++) {
				if (m >= b[i]) {
					sum += a[i];
					m = m - b[i];
				} else {
					sum += (float) m / b[i] * a[i];
					break;
				}
			}

		} else {
			System.out.println("兑换结束！！！！");
		}
		System.out.println(sum);

	}