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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.*;
import java.lang.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] A = new int[N+1];
		int[] B = new int[N+1];
		int[] r1 = new int[N+1];
		int[] r2 = new int[N+1];
		for(int i=1; i<=N; i++)
		{
			A[i]=sc.nextInt();
		}
		for(int i=1; i<=N; i++)
		{
			B[i]=sc.nextInt();
		}
		for(int i=1; i<=N; i++)
		{
			int count=0;
			for(int j=i+1; j<=N; j++)
			{
				if(A[j]<A[i])
				{
					count++;
				}
			}
			r1[i]=count;
		}
		for(int i=1; i<=N; i++)
		{
			int count=0;
			for(int j=i+1; j<=N; j++)
			{
				if(B[j]<B[i])
				{
					count++;
				}
			}
			r2[i]=count;
		}
		int sum=0;
		
		for(int i=1;i<=N; i++)
		{
			//System.out.println("r1: "+r1[i]);
		}
		
		for(int i=1; i<=N; i++)
		{
			//System.out.println("hii "+ (N-i));
			sum+=factorial(N-i)*r1[i];
			
		}
		int sum1=0;
		for(int i=1; i<=N; i++)
		{
			//System.out.println("byee");
			sum1+=factorial(N-i)*r2[i];
			
		}
		System.out.println(Math.abs(sum-sum1));
	}
	static int factorial(int n) 
    { 
        int res = 1, i; 
        for (i=2; i<=n; i++) 
            res *= i; 
        return res; 
    } 
	
}