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>
void mergeSort(int A[],int,int );
void merge(int A[],int,int,int );
int Mid;
int c=0;
int L[500000],R[500000];
int main(){
        int A[1000000];
        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++) {
                printf("%d",A[i]);
                if(i<n-1) printf(" ");
        }
        printf("\n");
        printf("%d\n",c);


        return 0;
}
void mergeSort(int A[],int Left,int Right){

        if(Left+1<Right) {
                int 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,j,l1,l2,k;
        l1=Mid-Left;
        l2=Right-Mid;

        for(i=0; i<l1; i++) {
                L[i]=A[Left+i];
        }
        for(j=0; j<l2; j++) {
                R[j]=A[Mid+j];
        }
        L[l1]=1000000000;
        R[l2]=1000000000;
        i=0;
        j=0;
        for(k=Left; k<Right; k++)
                if(L[i]<=R[j]) {
                        A[k]=L[i];
                        i++;
                        c++;

                }
                else{
                        A[k]=R[j];
                        j++;
                        c++;

                }
}