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<string.h>
#define max 1000000
void merge(int [],int ,int ,int );
void mergeSort(int [],int ,int );
int count = 0,newA[max];
int main(){
  int i,j,A[max],left,right,n;

  scanf("%d",&n);
  
  for(i = 0;i < n;i++){
    scanf("%d",&A[i]);
  }
  left = 0;
  right = n;
  mergeSort(A,left,right);
  
   for(i = 0;i < n-1;i++){
    printf("%d ",A[i]);
  }
   printf("%d\n",A[n-1]);
   printf("%d\n",count);
  return 0;
}

void mergeSort(int A[],int left, int right){
  int mid = 0;

  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 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] = A[left+i];
  }

  for(i = 0;i < n2;i++){
    R[i] = A[mid+i];
  }
  L[n1] = 10000000;
  R[n2] = 10000000;
  i = 0;
  j = 0;
  
  for(k = left;k < right;k++){
    if(L[i] <= R[j]){
     A[k] = L[i];
     i++;
    }
    else{
      A[k] = R[j];
      j++;
    }
    count++;
  }
}