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>
void merge(int *,int,int,int);
void mergeSort(int *,int,int);
int count=0;
int main(){
  int n,S[500000],i;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&S[i]);
  }
  mergeSort(S,0,n);

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

void mergeSort(int *A,int left,int right){
  int mid;
  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 i,j,k,n1,n2,*L,*R;
  n1=mid - left;
  n2=right - mid;

  L=malloc(n1*sizeof(L));
  R=malloc(n2*sizeof(R));
    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++){
    count++;
    if(L[i]<=R[j]){
      A[k] = L[i];
      i++;

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