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

int count=0;


int main(){

  int n,i;
  int s[500000];

  scanf("%d",&n);

  for(i = 0 ; i < n ; i++)
    scanf("%d",&s[i]);

  mergeSort(s,0,n);

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

  printf("%d\n",count);

  return 0;

}


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

void merge(int *s, int left, int mid, int right){

  int n1,n2,i,k,j;

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

  int l[n1+1],r[n2+1];

  for(i = 0 ; i <= n1-1 ; i++) l[i] = s[left + i];

  for(i = 0 ; i <= n2-1 ; i++) r[i] = s[mid + i];

  l[n1] = 1000000000;
  r[n2] = 1000000000;

  i = 0;
  j = 0;

  for(k = left ; k <= right-1 ; k++){

    count++;

    if(l[i] <= r[j]){ 
      s[k] = l[i];
      i = i+1;
    }
    else{
      s[k] = r[j];
      j = j+1;
    }
    
  }
}