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 MAX 500000
#define INFTY 20000000
 
void mergesort(int *,int,int);
void merge(int *,int,int,int);
 
int l[MAX/2+2],r[MAX/2+2];
int count;
int s[MAX];
 
int main()
{
 
  int n,i;
  count=0;
  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>0) printf(" ");
    printf("%d",s[i]);
  }
  printf("\n");
  printf("%d\n",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 i,j,k,n1,n2;
  n1=mid-left;
  n2=right-mid;
  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=i+1;
    }
    else {
      a[k]=r[j];
      j=j+1;
    }
  }
}