zoj3688棋盘多项式，有限制的排列问题

棋盘多项式本来的式子，但是这道题直接算出来棋盘多项式很麻烦，计算r_i的时候把问题转化为在1~2n的圆周上选取i个不相邻的数即可。1的时候特判一下。

#include 
#include 
#include 
#include 
using namespace std;
#define P 1000000007
#define N 200005
int n;
typedef long long LL;
LL fac[N],inv[N];
LL ans;
LL ksm(LL a, LL b)
{
	LL res = 1;
	while (b)
	{
		if (b & 1) res = (res * a) % P;
		a = a * a % P;
		b >>= 1;
	}
	return res;
}
void Fac()
{
	fac[0] = fac[1] = 1;
	for (LL i =1;i<=N-5;i++)
		fac[i] = fac[i-1] * i % P;
	inv[0] = inv[1] = 1;
	inv[N-5] = ksm(fac[N-5], P-2);
	for (int i=N-6;i>=2;i--)
		inv[i] = inv[i+1] * (i+1) % P;
}
LL r(int k)
{
	return 2*n % P * inv[k] % P * fac[n-k] % P * fac[2*n-k-1] % P * inv[2*n-2*k] % P;
}
int main()
{
	Fac();
	while (~scanf("%d", &n))
	{
		if (n == 1) {printf("%d\n", 0);continue;}
		ans = fac[n];
		for (int i=1;i<=n;i++)
		{
			if (i & 1)
				ans = (ans + P - r(i) % P) % P;
			else 
				ans = (ans + r(i) % P) % P;
		}
		printf("%lld\n", ans);
	}	
}