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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
int ans = 0;
int perm[9] = {};
int check[9] = {};
int flag = 0;
int permutation(int n, int p, int *arr) {
    if (n == p) {
        int cnt = 0;
        for (int i = 1; i < n; i++) {
            if (perm[i] == arr[i]) cnt++;
        }
        if (cnt == n - 1) flag = 1;
        if (flag) return 0;
        else return 1;
    }
    for (int i = 1 ; i < n; i++) {
        if (check[i]) continue;
        perm[p] = i;
        check[i] = 1;
        if(permutation(n, p+1, arr)) ans++;
        check[i] = 0;
    }
}

int main(void) {
    int n, idx, a, b;
    scanf("%d", &n);
    int P[n+1]; int Q[n+1];
    for (idx = 1; idx <= n; idx++) {
        scanf("%d", &P[idx]);
    }

    for (idx = 1; idx <= n; idx++) {
        scanf("%d", &Q[idx]);
    }

    permutation(n+1, 1, P);
    a = ans;
    ans = 0;
    flag = 0;

    for (idx = 1; idx <= n; idx++) {
        perm[idx] = 0;
        check[idx] = 0;
    }
    permutation(n+1, 1, Q);
    b = ans;
    printf("%d\n", abs(a-b));
    return 0;
}