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>
int C[1000000000];
int A[1000000000];
int count=0;
void merge(int a[],int left,int right);
int main(){
  int i,n;
 
  scanf("%d",&n);
  for(i=0; i < n ; i++){
    scanf("%d",&A[i]);
  }
  merge(A,0,n-1);
  for(i=0; i < n ; i++){
    if(i < n-1){
    printf("%d ",A[i]);
    }
    else{
      printf("%d",A[i]);
    }
  }
  printf("\n%d\n",count);
  return 0;
}

void merge(int A[],int left,int right){
  int mid,i,j,p;
  if(right <=left)return;
  mid=(left+right)/2;
  merge(A,left,mid);
  merge(A,mid+1,right);

  for(i=left; i <= mid;i++){
    C[i]=A[i];
  }
  j=right;
  for(i=mid+1;i <=right;i++){
    C[i]=A[j];
    j--;
  }
  j=right;
  i=left;
  for(p=left;p<=right;p++){
    if(C[i]<=C[j]){
      A[p]=C[i++];
      count++;
    }
    else{
      A[p]=C[j--];
      count++;
    }
  }
}