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.Collections;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		try (Scanner sc = new Scanner(System.in)) {

			int n = sc.nextInt();
			Integer[] p = new Integer[n];
			Integer[] q = new Integer[n];
			for(int i = 0 ; i < n ; i++) {
				p[i] = sc.nextInt();
			}
			for(int i = 0 ; i < n ; i++) {
				q[i] = sc.nextInt();
			}
			// 使える数字
			List<Integer> list1 = new LinkedList<>();
			List<Integer> list2 = new LinkedList<>();
			for(int i = 0 ; i < n ; i++) {
				list1.add(p[i]);
				list2.add(p[i]);
			}
			Collections.sort(list1);
			Collections.sort(list2);
			// その桁に何個あるか
			int[] tmp = new int[n];
			tmp[n - 1] = 1;
			tmp[n - 2] = 1;
			for(int i = n - 3 ; i >= 0 ; i--) {
				tmp[i] = tmp[i + 1] * (n - 1 - i);
			}
			int a = rank(p, tmp, list1);
			int b = rank(q, tmp, list2);
			System.out.println(Math.abs(a - b));

		}
	}

	private static int rank(Integer[] p, int[] tmp, List<Integer> list) {
		int rank = 1;
		for(int i = 0 ; i < p.length - 1 ; i++) {
			int a = p[i];
			int index = list.indexOf(a);
			rank += index * tmp[i];
			list.remove(index);
		}
		return rank;
	}
}