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 <stdlib.h>
#define N 500000
#define A 10000000


int res=0;

void mergesort(int *,int,int);
void merge(int *,int,int,int);

int main(){
  int i,n,S[N];

  scanf("%d",&n); 
  if(n>N){
    printf("error\n");
    exit(1);
  }

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

  mergesort(S,0,n);

  for(i=0 ; i<n-1 ;i++){
    printf("%d " ,S[i]);
  }
  printf("%d",S[n-1]);
  printf("\n%d\n",res);

  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=mid-left,n2=right-mid,i,j,k;
 int L[n1+1],R[n2+1];
  
  for(i=0 ; i<n1 ; i++){
    L[i]=S[left + i];
  }
  
  for(i=0 ; i<n2 ; i++){
    R[i]=S[mid + i];
  }
  
  L[n1]=A;
  R[n2]=A;
  i=0;
  j=0;
  
  for(k=left ; k<right ; k++){
    res++;
    if(L[i]<=R[j]){
      S[k]=L[i];
      i++;
    }
    else{
      S[k]=R[j];
      j++;
    }
    
  }
}