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: Wrong Answer)

import java.util.Scanner;

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 = findRank(p,n);
        int b = findRank(q,n);

        System.out.println((a - b) * -1);

    }

    public static int fac(int n) {
        int res = 1, i;
        for (i=2; i<=n; i++)
            res *= i;
        return res;
    }

    public static int findSmallerInRight(int[] array, int low, int high) {
        int countRight = 0, i;

        for (i = low + 1; i <= high; ++i)
            if (array[i] < array[low])
                ++countRight;

        return countRight;
    }

    public static int findRank(int[] array, int length) {
        int mul = fac(length);
        int rank = 1;
        int countRight;

        for (int i = 0; i < length; ++i) {
            mul /= length - i;

            // count number of chars smaller
            // than str[i] from str[i+1] to
            // str[len-1]
            countRight = findSmallerInRight(array, i, length - 1);

            rank += countRight * mul;
        }

        return rank;
    }
}