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: Accepted)

#include <stdio.h>
#include <stdlib.h>

#define N 500000
int count=0;

void Merge(int A[], int left, int mid, int right){
  int n1 = mid - left;
  int n2 = right - mid;
  int L[n1+1], R[n2+1];
  int i, j, k;

  for(i=0; i<n1; i++) L[i] = A[left + i];
  for(i=0; i<n2; i++) R[i] = A[mid + i];
  L[n1] = R[n2] = 1000000000;
  i = 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++;
    }
  }
}

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);
  }
}

int main(){
  int n, A[N], i;
  scanf("%d",&n);
  for(i=0; i<n; i++){
    scanf("%d",&A[i]);
  }
  MergeSort(A, 0, n);
  for(i=0; i<n-1; i++){
    printf("%d ",A[i]);
  }
  printf("%d\n",A[i]);
  printf("%d\n",count);
  return 0;
}