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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] array1 = new int[n];
        int[] array2 = new int[n];
        int[] used1 = new int[n];
        int[] used2 = new int[n];
        
        for (int i = 0; i < n; i++) {
            array1[i] = sc.nextInt() - 1;
        }
        for (int i = 0; i < n; i++) {
            array2[i] = sc.nextInt() - 1;
        }
        
        int count1 = 0;
        for (int i = 0; i < n-1; i++) {
            int tmp = 0;
            for (int j = 0; j <= array1[i]; j++) {
                if (used1[j] == 0) {
                    tmp++;
                }
            }
            count1 += tmp * fact(n - 1 - i);
            used1[array1[i]] = 1;
        }
        
        int count2 = 0;
        for (int i = 0; i < n-1; i++) {
            int tmp = 0;
            for (int j = 0; j <= array2[i]; j++) {
                if (used2[j] == 0) {
                    tmp++;
                }
            }
            count2 += tmp * fact(n - 1 - i);
            used2[array2[i]] = 1;
        }        
        System.out.println(Math.abs(count1 - count2));
    }
    
    public static int fact(int n) {
        int answer = 1;
        for (int i = 1; i <= n; i++) {
            answer = answer * i;
        }
        
        return answer;
    }
}