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>
#include<limits.h>

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

int num[500000];
int left=0,right;
int L[100000];
int R[100000];
int cnt;

int main(){
  int n,i;
  
  scanf("%d",&n);

  for(i=0;i<n;i++){
    scanf("%d",&num[i]);
  }
  right=n;
  cnt=0;

  mergeSort(left,right);

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

  return 0;
}

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

void merge(int left,int right,int mid){
  int i,j,k;
  int n1,n2;
  n1=mid-left;
  n2=right-mid;

  for(i=0;i<n1;i++){
    L[i]=num[left+i];
  }
  for(j=0;j<n2;j++){
    R[j]=num[mid+j];
  }
  
  L[n1]=10000;
  R[n2]=10000;

  i=0;
  j=0;
  
  for(k=left;k<right;k++){
    cnt++;
    if(L[i]<=R[j]){
      num[k]=L[i];
      i++;
    }

    else{
      num[k]=R[j];
      j++;
	}
  }
}