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 ENU 500000
#define INF 1000000000

void mergeSort(int S[], int, int);
void merge(int S[], int, int, int);
int left, right, S[ENU], enu, count = 0;


int main() {
	int i, j;

	scanf("%d", &enu);
	for (i = 0; i < enu; i++) {
		scanf("%d", &S[i]);
	}
	left = 0;
	right = enu;
	mergeSort(S, left, right);
	for (i = 0; i < enu - 1; i++) {
		printf("%d ", S[i]);
	}
	printf("%d\n", S[enu - 1]);
	printf("%d\n", count);
	return 0;
}


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

void merge(int S[], int left, int mid, int right) {
	int n1, n2, i, j, k;
	n1 = mid - left;
	n2 = right - mid;

	int L[n1], R[n2];
	for (i = 0; i < n1; i++) {
		L[i] = S[left + i];
	}
	for (j = 0; j < n2; j++) {
		R[j] = S[mid + j];
	}

	L[n1] = INF;
	R[n2] = INF;
	i = 0;
	j = 0;

	for (k = left; k < right; k++) {
		count++;
		if (L[i] <= R[j]) {
			S[k] = L[i];
			i++;
		}
		else {
			S[k] = R[j];
			j++;
		}
	}
}