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 <iostream>
using namespace std;
#define MAX 500000
#define SENTINEL 2000000000

int L[MAX/2+2],R[MAX/2+2];
int cnt;

void merge(int A[],int n,int left,int mid,int right)
{
	int n1 = mid - left;
	int n2 = right - mid;
	for(int i = 0; i < n1; i++) L[i] = A[left + i];
	for(int i = 0; i < n2; i++) R[i] = A[mid + i];
	L[n1] = R[n2] = SENTINEL;
	int i = 0, j = 0;
	for(int k = left; k < right; k++){
		cnt++;
		if(L[i] <= R[j]){
			A[k] = L[i++];
		}
		else{
			A[k] = R[j++];
		}
	}
}

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


int main()
{
	int A[MAX],n,i;
	cnt = 0;
	
	cin >> n;
	for(i = 0; i < n; i++) cin >> A[i];
	
	mergeSort(A,n,0,n);
	
	for(i = 0; i < n; i++){
		if(i) cout << " ";
		cout << A[i];
	}
	cout << endl;
	
	cout << cnt << endl;
	
	return 0;
}