description
给定(n),求$$f(n)=sum_{i=0}nsum_{j=0}i S(i,j) imes 2^j imes (j!)$$
data range
[1≤n≤100000
]
solution
第二类斯特林数及其递推式和展开式
(S(n,m))表示把(n)个有差别的球放入(m)个无差别的盒子中且无一空盒的方案数，递推式为
[S(n,m)=mS(n−1,m)+S(n−1,m−1),1≤m≤n−1
]
边界条件为:(S(n,n)=1(0≤n),S(n,0)=0(1≤n))
由容斥原理得其展开式为
[S(n,m)=frac{1}{m!}sum_{i=0}^{m}(-1)^iC_m^i(m-i)^n
]
考虑到当(i<j)时，(S(i,j)=0)，于是原式化为
[f(n)=sum_{i=0}^nsum_{j=0}^nsum_{k=0}^{j}(-1)^kC_j^k2^j(j-k)^i
]
千辛万苦得到了
[sum_{j=0}^n2^jj!sum_{k=0}^{j}frac{(-1)^k}{k!}frac{sum_{i=0}^{n}(j-k)^i}{(j-k)!}
]
不知道(sum_{i=0}^{n}(j-k)^i)怎么算?
等比数列啊!!!(sum_{i=0}^{n}(j-k)^i=frac{(j-k)^{n+1}-1}{j-k-1})
于是我们得到了最后的递推式
[sum_{j=0}^n2^jj!sum_{k=0}^{j}frac{(-1)^k}{k!}frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!}
]
继续(NTT)吧
code
注意(g(0)=1,g(1)=n+1)
#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const dd eps=1e-10;
const int mod=998244353;
const int N=1000010;
const dd pi=acos(-1);
const int inf=2147483647;
const ll INF=1e18+1;
il ll read(){
RG ll data=0,w=1;RG char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
return data*w;
}
il void file(){
srand(time(NULL)+rand());
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
}
il int poww(int a,int b){
RG int ret=1;
for(;b;b>>=1,a=1ll*a*a%mod)
if(b&1)ret=1ll*ret*a%mod;
return ret;
}
int l,r[N];
il void NTT(int *a,int n,int opt){
for(l=0;(1<<l)<n;l++);n=(1<<l);
for(RG int i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
for(RG int i=0;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
for(RG int i=2;i<=n;i<<=1){
RG int wn=poww(opt==1?3:(mod+1)/3,(mod-1)/i);
for(RG int j=0;j<n;j+=i){
RG int w=1;
for(RG int k=j;k<j+(i>>1);k++,w=1ll*w*wn%mod){
RG int x=1ll*a[k+(i>>1)]*w%mod;
a[k+(i>>1)]=(a[k]-x+mod)%mod;
a[k]=(a[k]+x)%mod;
}
}
}
if(opt==-1)
for(RG int i=0,rv=poww(n,mod-2);i<n;i++)
a[i]=1ll*a[i]*rv%mod;
}
int n,m,f[N],g[N],len,ans;
int pw[N],fac[N],inv[N];
il void init(){
pw[0]=fac[0]=inv[0]=1;
for(RG int i=1;i<=n;i++)
pw[i]=1ll*pw[i-1]*2%mod,fac[i]=1ll*fac[i-1]*i%mod;
inv[n]=poww(fac[n],mod-2);
for(RG int i=n-1;i;i--)inv[i]=1ll*inv[i+1]*(i+1)%mod;
}
int main()
{
n=read();init();g[0]=1;g[1]=n+1;
for(RG int i=0;i<=n;i++)f[i]=1ll*((i&1)?(mod-1):1)*inv[i]%mod;
for(RG int i=2;i<=n;i++)
g[i]=1ll*(poww(i,n+1)+mod-1)%mod*poww(i-1,mod-2)%mod*inv[i]%mod;
for(len=1;len<=(n<<1);len<<=1);
NTT(f,len,1);NTT(g,len,1);
for(RG int i=0;i<len;i++)f[i]=1ll*f[i]*g[i]%mod;
NTT(f,len,-1);
for(RG int i=0;i<=n;i++)(ans+=1ll*f[i]*pw[i]%mod*fac[i]%mod)%=mod;
printf("%d
",ans);return 0;
}