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 AAA 500000
#define BBB 1000000000

int count;
int L[AAA/2+2];
int R[AAA/2+2];

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


int main()
{
  int ori[AAA],n,i;
  
   scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&ori[i]);
  }

  mergeSort(ori,n,0,n);

  for(i=0;i<n-1;i++){
    printf("%d",ori[i]);
    printf("%c",' ');
  }

  printf("%d\n",ori[n-1]);
  printf("%d\n",count);

  return 0;
}

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

void merge(int ori[],int n,int left,int mid,int right)
{
  int i=mid-left;
  int j=right-mid;
  int k,l=0,m=0,o;

  for(k=0;k<i;k++) L[k]=ori[left+k];
  for(k=0;k<j;k++) R[k]=ori[mid+k];

  L[i]=R[j]=BBB;

  for(o=left;o<right;o++){
    count++;
    if(L[l]<=R[m]){
      ori[o]=L[l];
      l++;
    }
    else {
      ori[o]=R[m];
      m++;
    }
  }
}