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 P 250000
#define SENTINEL 100000000
int s[N];
long count=0;

void MergeSort(int *,int,int);
void Merge(int *,int,int,int);

main(){
  int n,i,left,right;

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

 
  left=0;
  right=n;

  MergeSort(s,left,right);

  for(i=0; i<n-1; i++)
    printf("%d ",s[i]);
  printf("%d\n%d\n",s[n-1],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,j,k;
  
  n1=mid-left;
  n2=right-mid;

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

  for(i=0; i<n1; i++)
    l[i]=s[left+i];
  for(i=0; i<n2; i++)
    r[i]=s[mid+i];
  
  l[n1]=SENTINEL;
  r[n2]=SENTINEL;
  
  i=0;
  j=0;
  
  for(k=left; k<right; k++){
    if(l[i]<=r[j]){
      s[k]=l[i];
      i++;
    }
    else {
      s[k]=r[j];
      j++;
    }
    count++;
  }
}