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(a,l,m,r)
  nl = m - l
  nr = r - m
  al = Array.new(nl+1)
  ar = Array.new(nr+1)
  for i in 0...nl
    al[i] = a[l+i]
  end
  for i in 0...nr
    ar[i] = a[m+i]
  end
  al[nl] = 1000000001
  ar[nr] = 1000000001
  i = 0
  j = 0
  cnt = 0
  for k in l...r
    if al[i] <= ar[j]
      a[k] = al[i]
      i += 1
    else
      a[k] = ar[j]
      j += 1
    end
    cnt += 1
  end
  cnt
end

def mergesort(a,l,r)
  cnt = 0
  if l+1 < r
    m = (l+r)/2
    cnt += mergesort(a,l,m)
    cnt += mergesort(a,m,r)
    cnt += merge(a,l,m,r)
  end
  cnt
end

n = gets.to_i
a = gets.split.map(&:to_i)
c = mergesort(a,0,n)
for i in 0...n
  print " " unless i == 0
  print a[i]
end
print "\n"
puts c