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: Wrong Answer)

#include <stdio.h>
#define N 500000
#define INFTY 9999999

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

int count = 0;
 
int main()
{
  int n,i,A[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,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,int left,int mid,int right)
{
  int n1 = mid - left;
  int n2 = right - mid;
  int i,j,k;
  int L[N],R[N];
 
  for(i = 0; i < n1; i++)
    L[i] = A[left + i];

  for(i = 0; i < n2; i++)
    R[i] = A[mid + i];

  L[n1] = INFTY;
  R[n2] = INFTY;

  i=0;
  j=0;

  for(k = left; k < right;k++)
    {
      count++;

    if(L[i] <= R[j])
      {
	A[k] = L[i];
	i++;
      }

    else
      {
	A[k] = R[j];
	j++;
      }
    }
}