Problem:
Merge Sort

Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.

Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1

Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)

Input

In the first line n is given. In the second line, n integers are given.

Output

In the first line, print the sequence S. Two consequtive elements should be separated by a space character.

In the second line, print the number of comparisons.

Constraints

n ≤ 500000

0 ≤ an element in S ≤ 109

Sample Input 1

10
8 5 9 2 6 3 7 1 10 4

Sample Output 1

1 2 3 4 5 6 7 8 9 10
34

Notes

Solution in Ruby (judge verdict: Accepted)

#!/usr/bin/ruby
$cnt = 0

def merge(a, left, mid, right)
  # mergeを指示された範囲を配列として取得
  # a[基準, 基準から何個とるか]
  l = a[left, (mid - left)]
  r = a[mid, (right - mid)]
  # loopが綺麗に回るように範囲超えの値を挿入
  l << 10**10
  r << 10**10
  # l, rの配列の先頭から小さい数字を詰めていく
  li, ri = 0, 0 # l, rの配列用のindex
  left.upto(right - 1) do | i | # aに挿入するためのindex
    $cnt += 1
    if l[li] <= r[ri]
      a[i] = l[li]
      li += 1
    else
      a[i] = r[ri]
      ri += 1
    end
  end
end

def merge_sort(a, left, right)
  if left + 1 < right
    mid = (left + right) / 2
    merge_sort(a, left, mid)
    merge_sort(a, mid, right)
    merge(a, left, mid, right)
  end
end

#
n = gets.to_i
a = gets.split.map(&:to_i)
merge_sort(a, 0, n)
puts a.join(' ')
puts $cnt