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>
#define M 500000
int a[M],b[M],count=0;
void mergeSort(int *a,int l,int r){
  int i,j,k,m;
  if(r>l){
    m=(r+l)/2;
    mergeSort(a,l,m);
    mergeSort(a,m+1,r);
    for(i=m+1;i>l;i--)
      b[i-1]=a[i-1];
    for(j=m;j<r;j++)
      b[r+m-j]=a[j+1];
    for(k=l;k<=r;k++){
      a[k]=(b[i]<b[j])?b[i++]:b[j--];
      count++;
    }
  }
}

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