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)

def merge(arr,left,mid,right)
  n1=mid-left
  n2=right-mid
  larr=Array.new()
  rarr=Array.new()
  0.upto(n1-1) do |i|
    larr[i]=arr[left+i]
  end
  0.upto(n2-1) do |i|
    rarr[i]=arr[mid+i]
  end
  larr[n1]=1000000000
  rarr[n2]=1000000000
  i=0
  j=0
  left.upto(right-1) do |k|
    if larr[i] <= rarr[j]
      arr[k]=larr[i]
      i=i+1
      @count+=1
    else
      arr[k]=rarr[j]
      j=j+1
      @count+=1
    end
  end
end

def mergeSort(arr,left,right)
  if left+1 < right
    mid=(left+right)/2
    mergeSort(arr,left,mid)
    mergeSort(arr,mid,right)
    merge(arr,left,mid,right)
  else
    return arr
  end
end

n=gets.to_i
arr=gets.split.map(&:to_i)
@count=0
mergeSort(arr,0,n)
puts arr.join(' ')
puts @count