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 C (judge verdict: Accepted)

#include<stdio.h>

void mergeSort(int *,int,int);
void merge(int *,int,int,int);

int A[500000],count=0;

int main()
{

  int i,n;

  scanf("%d",&n);
     
  for(i=0;i<n;i++)
    {
      scanf("%d",&A[i]);
    }
  
  mergeSort(A,0,n);

  for(i=0;i<n-1;i++)
    {
      printf("%d ",A[i]);
    }
  printf("%d\n%d\n",A[n-1],count);

  return 0;
}

void mergeSort(int A[500000], int left, int right)
{
  int mid;
  if (left+1 < right)
    {
      mid = (left + right)/2;
      mergeSort(A, left, mid);
      mergeSort(A, mid, right);
      merge(A, left, mid, right);
    }

}

void merge(int A[500000], int left, int mid, int right)
{
  int n1,n2,i,j,k;
  n1 = mid - left;
  n2 = right - mid;
  int L[n1+1],R[n2+1];
  for (i = 0;i<n1;i++)
    {
      L[i] = A[left + i];
    }
  for (i = 0;i<n2;i++)
    {
      R[i] = A[mid + i];
    }
  L[n1] = 2000000000;
  R[n2] = 2000000000;
  i = 0;
  j = 0;
  for (k = left;k<right;k++)
    {
      if (L[i] <= R[j])
	{
	  A[k] = L[i];
	  i = i + 1;
	  count++;
	}
      else
	{
	  A[k] = R[j];
	  j = j + 1;
	  count++;
	}
    }
}