Google Code Jam Notes-Rational Number Tree-Java

Problem: http://code.google.com/codejam/contest/2924486/dashboard#s=p1
Analysis: The trick is to map the Rational Number Tree to another complete binary tree as: 1 10 11 100 101 110 111You can see in this tree, left child is to append 0 the end of parent node, right child is to append 1 to the parent node. and each binary number value just represents the position :( 1, 10 (2), 11 (3), 100 (4), 101 (5 )..)
Then, since left child is p/(p + q), right child is (p + q)/q, so:
1. when p/q is given, if p> q, it is the left child (0), if p 2. when position is given. we can convert this value to binary representation, then we can generate the p/q step by step according to the complete binary tree we built.
Time complexity: O (log (n )).
My solution: (Your opinion is highly appreciated)

package codeJam.google.com;import java.io.BufferedReader;import java.io.FileReader;import java.io.FileWriter;import java.io.IOException;import java.math.BigInteger;/** * @author Zhenyi 2013 Dec 21, 2013 17:56:56 PM */public class RationalNumberTree {public static void main(String[] args) throws IOException {BufferedReader in = new BufferedReader(new FileReader("C:/Users/Zhenyi/Downloads/B-small-practice.in"));FileWriter out = new FileWriter("C:/Users/Zhenyi/Downloads/B-small-practice.out");// BufferedReader in = new BufferedReader(new FileReader(// "C:/Users/Zhenyi/Downloads/B-large-practice.in"));// FileWriter out = new FileWriter(// "C:/Users/Zhenyi/Downloads/B-large-practice.out");Integer T = new Integer(in.readLine());for (int cases = 1; cases <= T; cases++) {String[] st = in.readLine().split("\\s");Integer choice = new Integer(st[0]);if (choice.equals(1)) {// find p, qBigInteger n = new BigInteger(st[1]);BigInteger p = new BigInteger("1");BigInteger q = new BigInteger("1");int len = 0;BigInteger[] bits = new BigInteger[65];while (!n.equals(new BigInteger("0"))) {bits[len] = n.mod(new BigInteger("2"));n = n.divide(new BigInteger("2"));len++;}if (len == 1) {out.write("Case #" + cases + ": 1 1" + "\n");} else {for (int i = len - 2; i >= 0; i--) {if (bits[i].equals(new BigInteger("0"))) {q = q.add(p);}if (bits[i].equals(new BigInteger("1"))) {p = p.add(q);}}out.write("Case #" + cases + ": " + p + " " + q + "\n");}} else {// find sequenceBigInteger n = new BigInteger("0");BigInteger p = new BigInteger(st[1]);BigInteger q = new BigInteger(st[2]);BigInteger root = new BigInteger("1");BigInteger[] bits = new BigInteger[65];int len = 0;while (!(p.equals(root) && q.equals(root))) {if (p.subtract(q).signum() > 0) {// right childbits[len] = new BigInteger("1");p = p.subtract(q);} else {// left childbits[len] = new BigInteger("0");q = q.subtract(p);}len++;}bits[len] = new BigInteger("1");len++;if (len == 1) {out.write("Case #" + cases + ": 1" + "\n");} else {for (int i = len - 1; i >= 0; i--) {n = n.multiply(new BigInteger("2")).add(bits[i]);}out.write("Case #" + cases + ": " + n + "\n");}}}in.close();out.flush();out.close();}}