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 N 500001
int n,count;
void merge(int a[],int left,int mid,int right){
    int n1,n2,i,j,k;
    n1=mid-left;
    n2=right-mid;
    int l[n1+1],r[n2+1];
    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]=2000000000;
    i=j=0;
    for(k=left;k<right;k++){
        count++;
        if(l[i]<=r[j])a[k]=l[i++];
        else a[k]=r[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 i;
  int A[N];
 
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&A[i]);
  }
  mergeSort(A,0,n);
  for(i=0;i<n;i++){
      if(i) printf(" ");
      printf("%d",A[i]);
  }
  printf("\n");
  printf("%d\n",count);
  
  return 0;
}