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   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, q
				BigInteger 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 sequence
				BigInteger 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 child
						bits[len] = new BigInteger("1");
						p = p.subtract(q);

					} else {
						// left child
						bits[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();
	}
}

作者：u013301594 发表于2013-12-28 0:08:06 原文链接

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