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<math.h>

int A[500000], compare;

void merge(int left, int mid, int right){
  int n1 = mid - left, n2 = right - mid, L[n1 + 1], R[n2 + 1], 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] = INFINITY;
  R[n2] = INFINITY;
  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++;
    }
  compare++;
  }
}

void mergeSort(int left, int right){
  int mid;
  if(left + 1 < right){
    mid = (left + right)/2;
    mergeSort(left, mid);
    mergeSort(mid, right);
    merge(left, mid, right);
  }
}

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