外星人计算圆周率的java改版

/**
 * 采用展开的外星人算法求解PI
 * @author huxiao
 * 
 * Pi =2 + 1/3(2 +2/5(2 + ...(n/2n+1)(2+ ......
 *
 */


public class CalcPI { 
 public static final int MaxRound = 2000;
 public static int[] PiArray = new int[MaxRound];
 public static void CalcPi(int k)
 {
  int divide = 2*(MaxRound - k) + 3;  
  for(int i=1; i< MaxRound; i++ )
  {
   int relay = PiArray[i-1];
   PiArray[i-1] = relay / divide;
   PiArray[i] += (relay % divide) * 10000;
  }
  int multi = MaxRound - k + 1;
  
  for(int i = MaxRound-2; i>0; i--)
  {
   PiArray[i] *= multi;
  }
  for(int i = MaxRound-2; i>0; i--)
  {
   PiArray[i-1] += (PiArray[i])/10000;
   PiArray[i] = (PiArray[i])%10000;
  }  
  
  PiArray[0] += 2;
 }
 
 public static void main(String[] args){  
  PiArray[0] = 2;  
  for(int k=1; k<=MaxRound; k++)
  {
   CalcPi(k);
  }
  
  java.text.DecimalFormat format=new java.text.DecimalFormat("0000");

  for(int i=0; i<MaxRound-1; i++)
  {   
   System.out.print(format.format(PiArray[i]));
  }
 }
}