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>
#include <stdlib.h>
#include <math.h>

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

int *A,count=0;
int main()
{
  
  int n,i,left=0,mid,right;

  scanf("%d",&n);
  A=(int *)malloc(sizeof(int)*n);
  right=n;
  for(i=0;i<n;i++) scanf("%d",&A[i]);
  
  mergeSort(left,right);

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

  return 0;
}

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

void merge(int left,int mid,int right)
{
  int n1,n2,i,j,k;
  int *L,*R;

  n1=mid-left;
  n2=right-mid;

  L=(int *)malloc(sizeof(int)*n1+1);
  R=(int *)malloc(sizeof(int)*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]=(int)INFINITY;
  R[n2]=(int)INFINITY;
  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++;
	}
    }
}