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.io.*;
import java.util.*;

class Main{
    final static long MOD = 1000000007;
    FastScanner sc = new FastScanner();

    public static void main(String[] args) {
        new Main().solve();
    }

    void solve(){

        int n = sc.nextInt();
        //long n = sc.nextLong();
        String s = "";
        String s2 = "";
        String t = "";
        String t2 = "";
        for(int i = 0; i < n; i++){
            s += sc.next();
            t += i + 1;
            t2 += i + 1;
        }
        for(int i = 0; i < n; i++){
            s2 += sc.next();
        }


        int c = 0;
        int c2 = 0;
        while(t != null && !t.equals(s)){
            c++;
            t = nextPermutation(t);
        }
        while(t2 != null && !t2.equals(s2)){
            c2++;
            t2 = nextPermutation(t2);
        }

        System.out.println(Math.abs(c - c2));

    }

    String nextPermutation(String s){

        ArrayList<Character> list = new ArrayList<>();
        for(int i = 0; i < s.length(); i++) list.add(s.charAt(i));

        int pivotPos = -1;
        char pivot = 0;
        for(int i = list.size() - 2; i >= 0; i--){
            if(list.get(i) < list.get(i+1)){
    			pivotPos = i;
    			pivot = list.get(i);
    			break;
    		}
    	}

        if(pivotPos == -1 && pivot == 0) return null;

        int L = pivotPos + 1;
        int R = list.size() - 1;
    	int minPos = -1;
    	char min = Character.MAX_VALUE;
    	for(int i = R; i >= L; i--){
    		if(pivot < list.get(i)){
    			if(list.get(i) < min){
    				min = list.get(i);
    				minPos = i;
    			}
    		}
    	}

    	Collections.swap(list, pivotPos, minPos);
    	Collections.sort(list.subList(L, R + 1));

    	StringBuilder sb = new StringBuilder();
    	for(int i=0; i<list.size(); i++) sb.append(list.get(i));

    	return sb.toString();

    }

    long gcd(long a, long b){ // return aとbの最大公約数
        if(b == 0){
          return a;
        }
        return gcd(b, a % b);
    }

    long pow_mod(long a, long r){ // return a^r (mod MOD)
        long sum = 1;
        while(r > 0){
            if((r & 1) == 1){
                sum *= a;
                sum %= MOD;
            }
            a *= a;
            a %= MOD;
            r >>= 1;
        }
        return sum;
    }

    boolean is_prime(long a){
        if(a <= 1) return false;
        for(int i = 2; i*i <= a; i++){
            if(a % i == 0) return false;
        }
        return true;
    }

}



class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte(){
        if(ptr < buflen){
            return true;
        }else{
            ptr = 0;
            try{
                buflen = in.read(buffer);
            }catch(IOException e){
                e.printStackTrace();
            }
            if(buflen <= 0){
                return false;
            }
        }
        return true;
    }
    private int readByte(){ if(hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c){ return 33 <= c && c <= 126;}
    public boolean hasNext(){ while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next(){
        if(!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong(){
        if(!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        if(b < '0' || '9' < b){
            throw new NumberFormatException();
        }
        while(true){
            if('0' <= b && b <= '9'){
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt(){
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble(){ return Double.parseDouble(next());}
}