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: Wrong Answer)

#include <stdio.h>

#define N 500000

int cnt;

void merge(int* A, int left, int mid, int right) {
	int n1 = mid - left;
	int n2 = right - mid;
	int L[N], R[N];
	int i, j, k;

	for (i = 0; i <= n1 - 1; i++)
		L[i] = A[left + i];
	for (i = 0; i <= n2 - 1; i++)
		R[i] = A[mid + i];

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

	for (k = left; k <= right - 1; k++) {
		if (L[i] <= R[j]) {
			A[k] = L[i];
			i++;
		} else {
			A[k] = R[j];
			j++;
		}
	}
	cnt += (i + 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, n, S[N];

	scanf("%d", &n);
	for (i = 0; i < n; i++)
		scanf("%d", &S[i]);

	mergeSort(S, 0, n);

	for (i = 0; i < n; i++) {
		printf("%d", S[i]);
		if (i < n - 1)
			printf(" ");
	}
	printf("\n%d\n", cnt);

	return 0;
}