Problem:
Score : 300 points

Problem Statement

We have two permutations P and Q of size N (that is, P and Q are both rearrangements of (1,~2,~...,~N)).

There are N! possible permutations of size N. Among them, let P and Q be the a-th and b-th lexicographically smallest permutations, respectively. Find |a - b|.

Notes

For two sequences X and Y, X is said to be lexicographically smaller than Y if and only if there exists an integer k such that X_i = Y_i~(1 \leq i < k) and X_k < Y_k.

Constraints

2 \leq N \leq 8

P and Q are permutations of size N.

Input

Input is given from Standard Input in the following format:

N
P_1 P_2 ... P_N
Q_1 Q_2 ... Q_N

Output

Print |a - b|.

Sample Input 1

3
1 3 2
3 1 2

Sample Output 1

3

There are 6 permutations of size 3: (1,~2,~3), (1,~3,~2), (2,~1,~3), (2,~3,~1), (3,~1,~2), and (3,~2,~1). Among them, (1,~3,~2) and (3,~1,~2) come 2-nd and 5-th in lexicographical order, so the answer is |2 - 5| = 3.

Sample Input 2

8
7 3 5 4 2 1 6 8
3 8 2 5 4 6 7 1

Sample Output 2

17517

Sample Input 3

3
1 2 3
1 2 3

Sample Output 3

0

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

//atcoderはScannerクラスのsc.next()で行ける
public class Main {
    ArrayList<int[]> arrayList = new ArrayList<>();

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] p = new int[n];
        int[] q = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = sc.nextInt();
        }
        for (int k = 0; k < n; k++) {
            q[k] = sc.nextInt();
        }
        int[] data  = new int[n];
        for (int i = 1 ; i<=n ; i++){
            data[i-1]=i;
        }
        int cnt = 0;
        int a=0,b=0;
        do {
            cnt++;
            if (Arrays.equals(p,data)){
                a = cnt;
            }
            if (Arrays.equals(q,data)){
                b = cnt;
            }
        }
        while (findNextPermutation(data));
        System.out.println(Math.abs(a-b));

    }

    // Function to swap the data
    // present in the left and right indices
    public static int[] swap(int data[], int left, int right) {

        // Swap the data
        int temp = data[left];
        data[left] = data[right];
        data[right] = temp;

        // Return the updated array
        return data;
    }

    // Function to reverse the sub-array
    // starting from left to the right
    // both inclusive
    public static int[] reverse(int data[], int left, int right) {

        // Reverse the sub-array
        while (left < right) {
            int temp = data[left];
            data[left++] = data[right];
            data[right--] = temp;
        }

        // Return the updated array
        return data;
    }

    // Function to find the next permutation
    // of the given integer array
    public static boolean findNextPermutation(int data[]) {

        // If the given dataset is empty
        // or contains only one element
        // next_permutation is not possible
        if (data.length <= 1)
            return false;

        int last = data.length - 2;

        // find the longest non-increasing suffix
        // and find the pivot
        while (last >= 0) {
            if (data[last] < data[last + 1]) {
                break;
            }
            last--;
        }

        // If there is no increasing pair
        // there is no higher order permutation
        if (last < 0)
            return false;

        int nextGreater = data.length - 1;

        // Find the rightmost successor to the pivot
        for (int i = data.length - 1; i > last; i--) {
            if (data[i] > data[last]) {
                nextGreater = i;
                break;
            }
        }

        // Swap the successor and the pivot
        data = swap(data, nextGreater, last);

        // Reverse the suffix
        data = reverse(data, last + 1, data.length - 1);

        // Return true as the next_permutation is done
        return true;
    }
}