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>
#define N 500000
int tem[N];
int c=0;
void MG(int x[ ], int left, int right)
{
  int mid, i, j, k;
        
  if (left >= right)             
    return;                                                      
  mid = (left + right) / 2;      
  MG(x, left, mid);       
  MG(x, mid + 1, right);
  for (i = left; i <= mid; i++)
    tem[i] = x[i];     
  for (i = mid + 1, j = right; i <= right; i++, j--)
    tem[i] = x[j];
  i = left;        
  j = right;       
  for (k = left;k<= right; k++)   
    if (tem[i]<= tem[j]){        
      x[k]=tem[i++];c++;
    }
    else{
      x[k]=tem[j--];c++;
    }
}

int main(){
  int o1;
   int i,j;
   int A1[N];
 scanf("%d",&o1);
  for(i=0;i<o1;i++){
    scanf("%d",&A1[i]);
  }

  MG(A1,0,o1-1);
for (i = 0; i < o1; i++)
  {
    printf("%d",A1[i]);
 
 if(i!=o1 - 1)printf(" ");
  }
 printf("\n");
 printf("%d\n",c);
  return 0;
}