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)

# Your code here!

n = gets.chomp.to_i
a = gets.chomp.split.map(&:to_i)

@count = 0

def merge(a, l, m, r)
  l_arr = a[l, m-l] << 10000000000000000000
  l_i = 0
  r_arr = a[m, r-m] << 10000000000000000000
  r_i = 0
  
  (l...r).each do |idx|
    # 小さい方から入れていく
    if l_arr[l_i] <= r_arr[r_i]
      a[idx] = l_arr[l_i]
      l_i+=1
    else
      a[idx] = r_arr[r_i]
      r_i+=1
    end
    @count += 1
  end
end

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

merge_sort(a, 0, n)
puts a.map(&:to_s).join(' ')
p @count