codeforces 285 D Permutation Sum (状态压缩DP) 木

题目链接： http://www.codeforces.com/problemset/problem/285/D

Permutation p is an ordered set of integers p₁,  p₂,  ...,  p_n, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as p_i. We'll call number n the size or the length of permutation p₁,  p₂,  ...,  p_n.

Petya decided to introduce the sum operation on the set of permutations of length n. Let's assume that we are given two permutations of length n: a₁, a₂, ..., a_n and b₁, b₂, ..., b_n. Petya calls the sum of permutations a and b such permutation c of length n, where c_i = ((a_i - 1 + b_i - 1) mod n) + 1 (1 ≤ i ≤ n).

Operation means taking the remainder after dividing number x by number y.

Obviously, not for all permutations a and b exists permutation c that is sum of a and b. That's why Petya got sad and asked you to do the following: given n, count the number of such pairs of permutations a and b of length n, that exists permutation c that is sum of a and b. The pair of permutations x, y (x ≠ y) and the pair of permutations y, x are considered distinct pairs.

As the answer can be rather large, print the remainder after dividing it by 1000000007 (10⁹ + 7).

Input

The single line contains integer n (1 ≤ n ≤ 16).

Output

In the single line print a single non-negative integer — the number of such pairs of permutations a and b, that exists permutation c that is sum of a and b, modulo 1000000007 (10⁹ + 7).

Sample test(s)

Input

3

Output

18

Input

5

Output

1800

题目意思是： 求3个序列满足c[i] = (a[i]+b[i]-2)%n的个数， 每个序列都是n个不同的数字。 其实就是c[i] = (a[i]+b[i])%n（范围是0---n-1）
数据<=16，状态压缩的标志啊。
解法是对于一个给定的a序列，求满足条件的bc序列个数。用状态压缩求解。
依次检查b序列那些位没有用，而且c的(the position of a + the position of b)%n也没有用。则可以继续往下搜索。
求出的是a一个序列的解。而a有n！个序列，是不是每个都要求一次呢？ 很明显不是的。a的每个序列都是等价的，所以最终结果为：
ans*n!%mod. 其实这个算法还是很慢的，因为我跑n=15的时候跑了蛮久的。 预处理出来打表是可以的。

my python code：这个是状态压缩代码

ans = 0
mod = 1000000007
f = [1]*17
for i in xrange(1, 17):
    f[i] = (i*f[i-1])%mod
def dfs(Id, bNum, cSum, n):
    global ans
    if Id == n:
        ans += 1
        if ans >= mod : ans -= mod
        return 0    
    for i in xrange(n):
        if bNum&(1<<i): continue
        if cSum&(1<<( (Id+i)%n ) ): continue
        dfs(Id+1, bNum|(1<<i), cSum|(1<<( (Id+i)%n ) ), n )

if __name__ == '__main__':
    n = input()
    dfs(0, 0, 0, n)
    print ans*f[n]%mod

有了上面的代码就可以打表了：

num = [1, 18, 1800,670320,734832000,890786230,695720788,150347555]
n = input()
print 0 if not n&1 else num[n>>1]