求这个程序的算法！！在线等，急！！！！！！

[raingi520]

程序下载地址 http://ishare.iask.sina.com.cn/f/20514012.html

其中：
助火扇 是指 +2
取炭钳 是指 -2
小风箱 是指 *2

我想知道它的算法，能给出代码能使用的直接给分！！

[raingi520]

补充一下，这个程序的作用是，给出2个数，一个是A，另一个是B

A如何通过 +2、-2、*2 这三种运算操作，在12次运算（包含12次）以内得出B

[shine333]

public class Test {

  public static final String OPERATORS = " +-*";

  public static final int STEP_LIMIT = 12;

  public static final BigDecimal VALUE = BigDecimal.valueOf(2);

  public static void main(String[] args) throws Exception {
    int[] operators = new int[STEP_LIMIT];
    Scanner scanner = new Scanner(System.in);
    System.out.println("Please input A:");
    String aStr = scanner.next();
    BigDecimal a = new BigDecimal(aStr);
    System.out.println("Please input B:");
    String bStr = scanner.next();
    BigDecimal b = new BigDecimal(bStr);
    if (find(a, b, operators, 0)) {
      System.out.println(toExpression(a, b, operators));
    } else {
      System.out.println("Oops!");
    }
  }

  public static boolean find(BigDecimal from,
                             BigDecimal to,
                             int[] operators,
                             int step) throws Exception {
    if (step >= STEP_LIMIT) {
      return false;
    }
    if (from.compareTo(to) == 0) {
      return true;
    }
    for (int operator = 1; operator < OPERATORS.length(); operator++) {
      operators[step] = operator;
      if (find(newValue(from, operator), to, operators, step + 1)) {
        return true;
      }
    }
    operators[step] = 0;
    return false;
  }

  public static BigDecimal newValue(BigDecimal from, int operator) {
    switch (operator) {
      case 1:
        return from.add(VALUE);
      case 2:
        return from.subtract(VALUE);
      case 3:
        return from.multiply(VALUE);
      default:
        throw new RuntimeException();
    }
  }

  public static String toExpression(BigDecimal a, BigDecimal b, int[] operators) {
    StringBuilder buff = new StringBuilder(30).append(a);
    for (int i = 0; i < operators.length && operators[i] != 0; i++) {
      int operator = operators[i];
      buff.append(OPERATORS.charAt(operator)).append(VALUE);
    }
    return buff.append("=").append(b).toString();
  }
}

输出结果不考虑四则运算优先级。另外，不是最优解法。只是程序相对简单。