机修复模拟？

Question

问题陈述：

 A system is composed of 1, 2, or 3 machines and a repairman responsible for
 maintaining these machines. Normally, the machines are running and producing 
 a product. At random points in time, the machines fail and are fixed by the
 repairman. If a second or third machine fails while the repairman is busy
 fixing the first machine, these machines will wait on the services of the 
 repairman in a first come, first served order. When repair on a machine is 
 complete, the machine will begin running again and producing a product.
 The repairman will then repair the next machine waiting. When all machines
 are running, the repairman becomes idle.

 simulate this system for a fixed period of time and calculate the fraction
 of time the machines are busy (utilization) and the fraction of time
 the repairman is busy (utilization).

现在，输入是50运行时间和50修复时间，然后给出周期 计算它的利用率和要模拟的机器数量 对于每个测试用例。

示例输入：

7.0  4.5 13.0 10.5  3.0 12.0 ....
9.5  2.5  4.5 12.0  5.7  1.5 ....
20.0 1
20.0 3
0.0 0

示例输出：

       No of     Utilization
Case Machines Machine Repairman

 1       1       .525   .475
 2       3       .558   .775

案例2说明：

Machine Utilization   = ((7+3)+(4.5+6)+(13))/(3*20) = .558
Repairman Utilization =                   (15.5)/20 = .775

我的方法：

1）将机器加载到最小堆（称为runHeap）并给出每个堆    他们是一个运行时间，所以下一个给运行时间将是一个新的    从输入中的50个运行时间开始，

2）计算runHeap中最小提醒运行时间之间的最短时间    ，提醒维修队列头部的维修时间或提醒    时间来完成模拟，并称之为＃34; toGo＆＃34;。

3）通过toGo减去runHeap中所有机器的所有提醒运行时间，    通过toGo

减去repairQueue头部的提醒修复时间

4）所有提醒运行时间== 0的机器，将其推入repairQueue，    修复队列的头部如果提醒修复时间== 0将其推入    runHeap，

5）添加到当前时间

6）如果当前时间<1。模拟时间转到第2步，否则返回利用率。

现在，问题是一个好方法还是可以找出更好的方法？