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 INFTY 1000000000

int cnt;


void merge(int A[],int left,int mid,int right){
	int n1=mid-left;		//左側の要素数
	int n2=right-mid;		//右側の要素数
	
	int L[n1],R[n2];
	
	int i,j,k;
	for(i=0;i<n1;i++){		//左半分の要素代入
		L[i]=A[left+i];
	}
	for(i=0;i<n2;i++){		//右半分の要素代入
		R[i]=A[mid+i];
	}
	L[n1]=INFTY;
	R[n2]=INFTY;
	i=0;
	j=0;
	for(k=left;k<right;k++){
		cnt++;
		if(L[i]<=R[j]){
			A[k]=L[i];
			i++;
		}
		else{
			A[k]=R[j];
			j++;
		}
	}
}


void mergeSort(int A[],int left,int right){
	int mid;
	if(right-left>1){	//両端の交差の有無で判断
		mid=(left+right)/2;
		mergeSort(A,left,mid);
		mergeSort(A,mid,right);
		merge(A,left,mid,right);
	}
}


int main(void){
	int n;
	scanf("%d",&n);
	int i;
	int A[500000];
	cnt=0;
	for(i=0;i<n;i++){
		scanf("%d",&A[i]);
	}
	mergeSort(A,0,n);
	
	for(i=0;i<n;i++){
		if(i!=0){printf(" ");}
		printf("%d",A[i]);
	}
	
	printf("\n%d\n",cnt);
}