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>
#include<stdlib.h>
int *marge(int[], int, int);
int cnt;
int main(){
  int n, i;
  scanf("%d", &n);
  int *A;
  A = (int *)malloc(sizeof(int) * n);
  for(i = 0; i < n; i++)
    scanf("%d", &A[i]);
  cnt = 0;
  A = marge(A, 0, n);
  for(i = 0; i < n - 1; i++)
    printf("%d ", A[i]);
  printf("%d\n%d\n", A[i], cnt);
  free(A);
  return 0;
}

int *marge(int A[], int l, int r){
  if(l == r - 1){
    int *tmp;
    tmp = (int *)malloc(sizeof(int));
    tmp[0] = A[l];
    return tmp;
  }
  int *L, *R;
  L = (int *)malloc(sizeof(int) * ((r - l) / 2));
  R = (int *)malloc(sizeof(int) * ((r - l + 1) / 2));
  L = marge(A, l, (l + r) / 2);
  R = marge(A, (l + r) / 2, r);
  int *B;
  int i, j, k;
  i = j = k = 0;
  B = (int *)malloc(sizeof(int) * (r - l));
  while(1){
    if(i < (r - l) / 2 & j < (r - l + 1) / 2){
      if(L[i] < R[j]){
	B[k] = L[i];
	i++;
	k++;
      }
      else{
	B[k] = R[j];
	j++;
	k++;
      }
      cnt++;
    }
    else{
      if(i < (r - l) / 2 | j < (r - l + 1) / 2){
	if(i < (r - l) / 2){
	  B[k] = L[i];
	  i++;
	  k++;
	}
	else{
	  B[k] = R[j];
	  j++;
	  k++;
	}
	cnt++;
      }
      else
	break;
    }
  }
  free(L);
  free(R);
  return B;
}