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 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 i, j, k;
  int n1 = mid -left;
  int n2 = right - mid;
  for(i=0;i<n1;i++) L[i] = A[left + i];
  for(i=0;i<n2;i++) R[i] = A[mid + i];
  L[n1] = R[n2] = SENTINEL;
  i=0, j=0;
  for(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) {
  int mid;
  if(left+1 < right) {
    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;
  cnt = 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++) {
    if(i) printf(" ");
    printf("%d",A[i]);
  }
  printf("\n%d\n",cnt);

  return 0;
}