【BZOJ4555】【TJOI2016】【HEOI2016】求和 第二类斯特林数 NTT

题目大意

  求\(f(n)=\sum_{i=0}^n\sum_{j=0}^i2^j\times j!\times S(i,j)\\\)

  对\(998244353\)取模
  
  \(n\leq 100000\)

题解

\[\begin{align} S(n,k)&=\frac1{k!}\sum_{i=0}^k{(-1)}^i\binom{k}{i}{(k-i)}^n\\ &=\frac1{k!}\sum_{i=0}^k{(-1)}^i\frac{k!}{i!(k-i)!}(k-i)^n\\ &=\sum_{i=0}^k\frac{{(-1)}^i}{i!}\frac{{(k-i)}^n}{(k-i)!} \end{align} \]

  因为\(S(i,j)=0~(i<j)\),所以

\[\begin{align} f(n)&=\sum_{i=0}^n\sum_{j=0}^n2^j\times j!\times S(i,j)\\ &=\sum_{j=0}^n2^j\times j!\times\sum_{i=0}^nS(i,j)\\ &=\sum_{j=0}^n2^j\times j!\times\sum_{i=0}^n\sum_{l=0}^j\frac{{(-1)}^i}{l!}\frac{{(j-l)}^i}{(j-l)!}\\ &=\sum_{j=0}^n2^j\times j!\times\sum_{l=0}^j\frac{{(-1)}^i}{l!}\sum_{i=0}^n\frac{{(j-l)}^i}{(j-l)!} \end{align} \]

  设

\[A(x)=\frac{{(-1)}^i}{i!},B(x)=\sum_{i=0}^n\frac{x^i}{x!} \]

  所以

\[B(x)=\frac{x^{n+1}-1}{x!(x-1)} \]

\[f(n)=\sum_{j=0}^n2^j\times j!\times \sum_{i=0}^jA(i)B(j-i) \]

  直接上NTT

  时间复杂度:\(O(n\log n)\)

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
ll p=998244353;
ll fp(ll a,ll b)
{
	ll s=1;
	while(b)
	{
		if(b&1)
			s=s*a%p;
		a=a*a%p;
		b>>=1;
	}
	return s;
}
namespace ntt
{
	ll w1[1000010];
	ll w2[1000010];
	int rev[1000010];
	int n;
	void init()
	{
		n=262144;
		int i;
		for(i=2;i<=n;i<<=1)
		{
			w1[i]=fp(3,(p-1)/i);
			w2[i]=fp(w1[i],p-2);
		}
		rev[0]=0;
		for(i=1;i<n;i++)
			rev[i]=(rev[i>>1]>>1)|(i&1?n>>1:0);
	}
	void ntt(ll *a,int t)
	{
		int i,j,k;
		ll w,wn,u,v;
		for(i=0;i<n;i++)
			if(rev[i]<i)
				swap(a[i],a[rev[i]]);
		for(i=2;i<=n;i<<=1)
		{
			wn=(t==1?w1[i]:w2[i]);
			for(j=0;j<n;j+=i)
			{
				w=1;
				for(k=j;k<j+i/2;k++)
				{
					u=a[k];
					v=a[k+i/2]*w%p;
					a[k]=(u+v)%p;
					a[k+i/2]=(u-v+p)%p;
					w=w*wn%p;
				}
			}
		}
		if(t==-1)
		{
			ll inv=fp(n,p-2);
			for(i=0;i<n;i++)
				a[i]=a[i]*inv%p;
		}
	}
};
ll a[500010];
ll b[500010];
ll fac[200010];
int main()
{
	ntt::init();
	int n;
	scanf("%d",&n);
	int i;
	fac[0]=1;
	for(i=1;i<=n;i++)
		fac[i]=fac[i-1]*i%p;
	a[0]=1;
	for(i=1;i<=n;i++)
		a[i]=(((i&1?-1:1)*fp(fac[i],p-2))%p+p)%p;
	b[0]=1;
	b[1]=n+1;
	for(i=2;i<=n;i++)
		b[i]=((fp(i,n+1)-1)*fp(fac[i]*(i-1)%p,p-2)%p+p)%p;
	ntt::ntt(a,1);
	ntt::ntt(b,1);
	for(i=0;i<ntt::n;i++)
		a[i]=a[i]*b[i]%p;
	ntt::ntt(a,-1);
	ll ans=0;
	for(i=0;i<=n;i++)
		ans=(ans+fp(2,i)*fac[i]%p*a[i]%p)%p;
	printf("%lld\n",ans);
	return 0;
}
posted @ 2018-03-05 20:21  ywwyww  阅读(158)  评论(0编辑  收藏  举报