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 MAX 500000
#define INFTY 2000000000


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

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] = INFTY;
	int i = 0,j = 0;
	for(int 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 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(void){
	int A[MAX],n,i;
	count = 0;

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

}