Problem:
Score : 400 points

Problem Statement

Caracal is fighting with a monster.

The health of the monster is H.

Caracal can attack by choosing one monster. When a monster is attacked, depending on that monster's health, the following happens:

If the monster's health is 1, it drops to 0.

If the monster's health, X, is greater than 1, that monster disappears. Then, two new monsters appear, each with the health of \lfloor X/2 \rfloor.

(\lfloor r \rfloor denotes the greatest integer not exceeding r.)

Caracal wins when the healths of all existing monsters become 0 or below.

Find the minimum number of attacks Caracal needs to make before winning.

Constraints

1 \leq H \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H

Output

Find the minimum number of attacks Caracal needs to make before winning.

Sample Input 1

2

Sample Output 1

3

When Caracal attacks the initial monster, it disappears, and two monsters appear, each with the health of 1.

Then, Caracal can attack each of these new monsters once and win with a total of three attacks.

Sample Input 2

4

Sample Output 2

7

Sample Input 3

1000000000000

Sample Output 3

1099511627775

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long h = getLong(sc);
		List<Long> monsters = new ArrayList<>();
		monsters.add(h);

		long count = calcAttack(h, 1, 0);

		System.out.println(count);
	}

	private static long calcAttack(long hp, long monsters, long count) {
		count += monsters;
		if (hp == 1) {
			return count;
		} else {
			return calcAttack(hp / 2, monsters * 2, count);
		}

	}

	private static boolean isEqual(Integer[] perm, int[] target) {
		for (int i = 0; i < perm.length; i++) {

			if (perm[i].intValue() != target[i]) {

				return false;
			}
		}
		return true;
	}

	/**
	 * *汎用メソッド<br>
	 * 入力した配列に対する順列を取得
	 *
	 * @param numArray
	 * @return
	 */
	private static List<Integer[]> getPermutation(int[] numArray) {
		List<Integer[]> permutation = new ArrayList<>();
		Integer[] ans = new Integer[numArray.length];
		boolean[] used = new boolean[numArray.length];

		return buildPermutaion(permutation, numArray, ans, used, 0);
	}

	/**
	 * *汎用メソッド(内部)<br>
	 * 入力した配列に対する順列を作成
	 *
	 * @param list
	 * @param numArray
	 * @param ans
	 * @param used
	 * @param depth
	 * @return
	 */
	private static List<Integer[]> buildPermutaion(List<Integer[]> list, int[] numArray, Integer[] ans, boolean[] used,
			int depth) {

		// 全部使い切った時=>できた順列を返す
		if (depth == numArray.length) {
			list.add(ans.clone());
			return list;
		}
		// 使い切っていない場合

		for (int i = 0; i < numArray.length; i++) {
			if (used[i]) {
				continue;
			}
			ans[depth] = numArray[i];
			used[i] = true;
			list = buildPermutaion(list, numArray, ans, used, depth + 1);

			used[i] = false;
		}

		return list;
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をlongで返却
	 *
	 * @return
	 */
	public static long getLong(Scanner sc) {

		return Long.parseLong(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をintで返却
	 *
	 * @return
	 */
	public static int getInt(Scanner sc) {

		return Integer.parseInt(sc.next());
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をStringで返却
	 *
	 * @return
	 */
	public static String getString(Scanner sc) {

		return sc.next();
	}

	/**
	 * *汎用メソッド <br>
	 * スキャナーからの入力値をint配列で返却
	 *
	 * @return
	 */
	public static int[] getIntArray(Scanner sc) {
		String line = "";
		// 直前にnext()を実行していると改行コードだけ取ってしまうのでその対策
		while (line.equals("")) {
			line = sc.nextLine();
		}

		String[] strArray = line.split(" ");
		int[] intArray = new int[strArray.length];
		for (int i = 0; i < strArray.length; i++) {
			intArray[i] = Integer.parseInt(strArray[i]);
		}
		return intArray;
	}

}