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>
 
#define MAX 500000
#define INFTY 1000000000
 
void MargeSort(int *,int,int);
void Marge(int *,int,int,int);
 
int count=0;
 
int main(void){
  int n,i,A[MAX];
 
  scanf("%d", &n);
  for(i=0;i<n;i++) scanf("%d", &A[i]);
 
  MargeSort(A,0,n);
  //for(i=0;i<n;i++) printf("%d", A[i]);
  //printf("\n");
 
  for(i=0;i<n;i++){
    if(i!=n-1) printf("%d ",A[i]);
    else printf("%d\n", A[i]);
  }
  printf("%d\n", count);
 
  return 0;
}
 
void MargeSort(int *A,int left, int right){
  int mid;
 
  if(left+1<right){
    mid=(left+right)/2;
    MargeSort(A,left,mid);
    MargeSort(A,mid,right);
    Marge(A,left,mid,right);
  }
 
}
 
void Marge(int *A,int left, int mid,int right){
  int n1=mid-left;
  int n2=right-mid;
  int L[n1+1],R[n2+1];
  int i,j,k;
 
  for(i=0;i<n1;i++) L[i]=A[left+i];
  for(i=0;i<n2;i++) R[i]=A[mid+i];
 
  L[n1]=INFTY; R[n2]=INFTY; i=0; j=0;
 
  for(k=left;k<right;k++){
    if(L[i]<=R[j]){
      A[k]=L[i];
      i++;
    } else {
      A[k]=R[j];
      j++;
    }
    count++;
  }
 
}