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>
#define INFINITY 100000000

int A[500000];
int num=0;
int n1,n2;

void merge(int A[],int left,int mid,int right){
  int i,k,j;
  n1=mid-left;
  n2=right-mid;
  float L[n1],R[n2];
  for(i=0;i<n1;i++)L[i]=A[left+i];
  for(i=0;i<n2;i++)R[i]=A[mid+i];
  L[n1]=INFINITY;
  R[n2]=INFINITY;
  i=0;
  j=0;
  for(k=left;k<right;k++){
    if(L[i]<=R[j]){
      A[k]=L[i];
      i=i+1;
num++;
    }
   else{
     num++;
  A[k]=R[j];
  j=j+1;
    }
  }
}
void mergesort(int A[],int left,int right){
  if(left+1<right){
    int mid;
    mid = (left + right)/2;
    mergesort(A, left, mid);
    mergesort(A, mid, right);
    merge(A, left, mid, right);
  }
}
int main(){
  int n,i;
  scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d",&A[i]);
}
mergesort(A,0,n);
for(i=0;i<n;i++){
if(i==n-1){
printf("%d\n",A[i]);
break;
}
printf("%d ",A[i]);
}
printf("%d\n",num);

return 0;
}