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  MID  500000
#define   N   1000000
#define INFTY 2000000000

void mergeSort(int*,int,int);
void merge(int*,int,int,int);

int count=0;

int main(){
  int i;
  int n;
  int A[N];

  scanf("%d",&n);

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

  mergeSort(A,0,n);

  for(i=0;i<n-1;i++){
  printf("%d ",A[i]);
  }
  printf("%d\n",A[i]);
  printf("%d\n",count);
  
  return 0;
}

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);
  }
}

void merge(int *A,int left,int mid,int right){

  int i;
  int j;
  int k;
  int n1 = mid - left;
  int n2 = right - mid;
  int L[MID];
  int R[MID];
  
  for(i=0;i<n1;i++){
    L[i] = A[left+i];
  }
  
  for(j=0;j<n2;j++){
    R[j] = A[mid+j];
  }
    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 = i + 1;
    }
    else{ 
      A[k] = R[j];
      j = j + 1;
    }
    count++;
  }
}