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: Wrong Answer)


#include <stdio.h>
void mergesort( int [],int,int);
void merge(int [],int,int,int);

int c =0, n ;

int main()
{
  int i ;
  scanf("%d", &n);
  int s[n];
  for( i = 0; i < n; i++){
    scanf("%d" , &s[i]);
  }
  mergesort(s , 0 ,n );
  for ( i = 0; i < n; i++){
    printf("%d", s[i]);
    if ( i != n-1){printf(" ");}
  }
  printf("\n%d\n",c);
  return 0;
}
void mergesort( int a[] ,int lef,int rig){
    
  if ( lef +1 < rig){
    int mid = ( lef + rig)/2;
    mergesort( a , lef ,mid);
    mergesort(a , mid ,rig);
    merge (a ,lef ,mid ,rig);
  }
}
void merge ( int a[] , int lef , int mid ,int rig){
  int k,none = mid - lef,i,j,	ntwo = rig - mid, l[none ], r[ ntwo];
  for (i = 0; i < none;i++ ){
    l[i] = a[lef+ i];
  }
  for ( i = 0; i < ntwo ; i++){
    r[i] = a[mid + i];
  }l[none] = 100000001;r[ntwo]= 1000000001; i= 0;j= 0;
  for (k = lef ;k < rig ;k++ ){
    c++;
    if (l[i] <= r[j]){
      a[k] = l[i];
      i = i + 1;
    }else{ 
      a[k] = r[j];
      j = j + 1;
    }
  }
}