[ZJOI2010]排列计数

\([ZJOI2010]\)排列计数

链接:https://www.luogu.com.cn/problem/P2606

题面:求满足\(p_{i}>p_{i/2}(∀i∈[2,n])\)的排列个数。

题解:考虑将限制转化成一棵树,如果我们将\(i\)向\(i/2\)连边,这道题目变成了树上排列计数的裸题。转化为了一个树后,限制便变为了儿子的权值大于父亲,令\(dp_{i}\)表示\(i\)的子树的排列个数,转移时需要用多重集的排列数合并,则

\(dp_{i}=(sz_{x}-1)!\times\dfrac{1}{\prod_{t∈son(i)}^{}sz_{t}!}\prod_{t∈son(i)}^{}dp_{t}\)

#include<iostream>
#include<cstdio>
using namespace std;
long long sz[1000001],n,m,dp[1000001],fac[1000001],invfac[1000001];
void dfs(int x)
{
	sz[x]=1;
	dp[x]=1;
	if (x*2<=n)
	{
		dfs(x*2);
		dp[x]=dp[x]*dp[x*2]%m;
		dp[x]=dp[x]*invfac[sz[x*2]]%m;
		sz[x]+=sz[x*2];
	}
	if (x*2+1<=n)
	{
		dfs(x*2+1);
		dp[x]=dp[x]*dp[x*2+1]%m;
		dp[x]=dp[x]*invfac[sz[x*2+1]]%m;
		sz[x]+=sz[x*2+1];
	}
	dp[x]=dp[x]*fac[sz[x]-1]%m;
	return;
}
long long fast_pow(long long a,int b)
{
	if (b==0)
		return 1;
	if (b&1)
		return fast_pow(a*a%m,b/2)*a%m;
	else
		return fast_pow(a*a%m,b/2); 
}
int main()
{
	cin>>n>>m;
	fac[0]=1;
	for (int i=1;i<=n;++i)
		fac[i]=fac[i-1]*i%m;
	invfac[n]=fast_pow(fac[n],m-2);
	for (int i=n-1;i>=0;--i)
		invfac[i]=invfac[i+1]*(i+1)%m;
	dfs(1);
	cout<<dp[1]<<endl;
	return 0;
}