Google Code Jam Notes - Rational Number Tree - Java

Problem:

Retrieved from: http://code.google.com/codejam/contest/2924486/dashboard#s=p1

Consider an infinite complete binary tree where the root node is 1/1 and left and right childs of node p/q are p/(p+q) and (p+q)/q, respectively. This tree looks like:

         1/1    ______|______    |           |   1/2         2/1 ___|___     ___|___ |     |     |     |1/3   3/2   2/3   3/1...

It is known that every positive rational number appears exactly once in this tree. A level-order traversal of the tree results in the following array:

1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, ...

Please solve the following two questions:

1. Find the n-th element of the array, where n starts from 1. For example, for the input 2, the correct output is 1/2.
2. Given p/q, find its position in the array. As an example, the input 1/2 results in the output 2.

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of one line. The line contains a problem id (1 or 2) and one or two additional integers:

1. If the problem id is 1, then only one integer n is given, and you are expected to find the n-th element of the array.
2. If the problem id is 2, then two integers p and q are given, and you are expected to find the position of p/q in the array.

Output

For each test case:

1. If the problem id is 1, then output one line containing "Case #x: p q", where x is the case number (starting from 1), and p, q are numerator and denominator of the asked array element, respectively.
2. If the problem id is 2, then output one line containing "Case #x: n", where x is the case number (starting from 1), and n is the position of the given number.

Limits

1 ≤ T ≤ 100; p and q are relatively prime.

Small dataset

1 ≤ n, p, q ≤ 2¹⁶-1; p/q is an element in a tree with level number ≤ 16.

Large dataset

1 ≤ n, p, q ≤ 2⁶⁴-1; p/q is an element in a tree with level number ≤ 64.

Sample

Input 
 
Output 
 

41 22 1 21 52 3 2

Case #1: 1 2Case #2: 2Case #3: 3 2Case #4: 5

Analysis:

The trick is to map the Rational Number Tree to another complete binary tree as:

                         1

                     10   11

            100  101   110   111

You 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<q, it is the right child (1). Also we can calculate their parent's value. By calculating this repeatedly, until we get the root child 1/1, we can get a successive binary numbers, their value is the position.

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();}}