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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int asc(const void* a, const void* b) {
    return *(int*)a - *(int*)b;
}
int main() {
    int a,n,order_p,order_q;
    int p[8] = { 0 }, q[8] = { 0 };
    int upper_new = 1, lower_new = 1;
    int upper_old = 1, lower_old = 1;
    int flag_p[9] = { 10,10,10,10,10,10,10,10,10 }, flag_q[9] = { 10,10,10,10,10,10,10,10,10 };

    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &p[i] );
    for (int i = 0; i < n; i++) scanf("%d", &q[i]);
    for (int i = 1; i <= n; i++) {
        upper_old *= i;
        flag_p[i - 1] = i;
        flag_q[i - 1] = i;
    }
    //printf("upper_old=%d\n", upper_old);
    for (int i = 0; i < n - 1; i++) {
        qsort(flag_p, n, sizeof(int), asc);
        for( a=0;a < n - i; a++ ){
            if (p[i] == flag_p[a]) {
                flag_p[a] = 10;
                break;
            }
        }
        //printf("in pa=%d\n", a);
        upper_new = (upper_old - lower_old + 1) * (a + 1) /(n - i) + lower_old - 1;
        lower_new = (upper_old - lower_old + 1) * a /( n - i) + lower_old;
        upper_old = upper_new;
        lower_old = lower_new;
    }
    order_p = upper_new;

    upper_new = 1, lower_new = 1;
    upper_old = 1, lower_old = 1;
    for (int i = 1; i <= n; i++) upper_old *= i;
    for (int i = 0; i < n - 1; i++) {
        qsort(flag_q, n, sizeof(int), asc);
        for (a = 0; a < n - i; a++) {
            if (q[i] == flag_q[a]) {
                flag_q[a] = 10;
                break;
            }
        }
        upper_new = (upper_old - lower_old + 1) * (a + 1) / (n - i) + lower_old - 1;
        lower_new = (upper_old - lower_old + 1) * a / (n - i) + lower_old;
        upper_old = upper_new;
        lower_old = lower_new;
    }
    order_q = upper_new;
    //printf("order_p=%d\norder_q=%d\n", order_p, order_q);
    printf("%d", abs(order_p-order_q));
    return 0;
}