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) {
        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 i = 0; i < n; i++) q[i] = sc.nextInt();
        int a = func(p);
        int b = func(q);
        System.out.println(Math.abs(a - b));
    }
    public static int func(int[] p) {
        int ret = 0;
        int n = p.length;
        boolean[] used = new boolean[n];
        for (int i = 0; i < n; i++) {
            int cnt = 0;
            for (int j = 1; j < p[i]; j++) {
                if (!used[j - 1]) cnt++;
            }
            ret += cnt * frac(n - 1 - i);
            used[p[i] - 1] = true;
        }
        return ret;
    }
    public static int frac(int x) {
        if (x == 0) return 1;
        return x * frac(x - 1);
    }
}