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

 程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=500005;
const int INF=0x3f3f3f3f;
int n;
int a[maxn];
int cnt=0;
void Sort (int left,int mid,int right)
{
    int L[(maxn>>1)+10],R[(maxn>>1)+10];
    int numl=mid-left;
    int numr=right-mid;
    for (int i=0;i<numl;i++)
        L[i]=a[i+left];
    for (int i=0;i<numr;i++)
        R[i]=a[i+mid];
    L[numl]=R[numr]=INF;
    for (int i=left,j=0,k=0;i<right;i++)
    {
        cnt++;
       a[i]=L[j]<R[k]?L[j++]:R[k++];
    }
}
//先進行分治，當分治到不能分治的時候進行回溯排序
void Merge (int left,int right)
{
    if(left+1<right)
    {
        int mid=(left+right)>>1;
        Merge (left,mid);
        Merge (mid,right);
        Sort (left,mid,right);
    }
}
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
        scanf("%d",&a[i]);
    Merge (0,n);
    for (int i=0;i<n;i++)
        printf("%d%c",a[i],i==n-1?'\n':' ');
    printf("%d\n",cnt);
    return 0;
}