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 INFTY 10000000

void merge(int *,int,int,int);
void mergeSort(int *,int,int);
  
int L[5000000],R[5000000],count=0;

int main() {

  int n,i,S[5000000];
  
  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) printf(" ");
    printf("%d",S[i]);
  }
  printf("\n%d\n",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;

  for(i = 0; i < n1; i++) {
    L[i] = S[left + i];
  }
  for(i = 0; i < n2; i++) {
    R[i] = S[mid + i];
  }
  L[n1] = INFTY;
  R[n2] = INFTY;
  i = 0;
  j = 0;
  
  for(k = left; k < right; k++) {
    count++;
    if(L[i] <= R[j]) {
      S[k] = L[i];
      i = i + 1;
    }
    else {
      S[k] = R[j];
      j = j + 1;
    }
}
}