bzoj4555(斯特林数)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=4555

题解

还是用 $S_2$ 的卷积展开

然后做个卷积就可以了。。

等比数列求和那里需要特判 $0$ 和 $1$ 的情况。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<complex>
#include<cstdlib>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 270005
#define nm 100005
const double pi=acos(-1);
const ll inf=998244353;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}


int n;
ll a[NM],b[NM],inv[NM],invp[NM],p[NM],ans;

int lim,bit,rev[NM],W[NM],w[NM],invn;

inline void reduce(ll&x){x+=x>>63&inf;}
ll qpow(ll x,ll t){
ll s=1;
for(;t;t>>=1,x=x*x%inf)if(t&1)s=s*x%inf;
return s;
}
void init(int m){
for(lim=1,bit=0;lim<m;lim<<=1)bit++;invn=qpow(lim,inf-2);
inc(i,0,lim-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
ll t=qpow(3,(inf-1)/lim);W[0]=1;
inc(i,1,lim)W[i]=W[i-1]*t%inf;
}
void fft(ll*a,int f=0){
int n=lim;
inc(i,1,n-1)if(i<rev[i])swap(a[i],a[rev[i]]);
for(int k=1;k<n;k<<=1){
int t=n/k>>1;
for(int i=0,j=0;i<k;i++,j+=t)w[i]=W[f?n-j:j];
for(int i=0;i<n;i+=k<<1)
for(int j=0;j<k;j++){
ll x=a[i+j],y=w[j]*a[i+j+k]%inf;
reduce(a[i+j]=x+y-inf);reduce(a[i+j+k]=x-y);
}
}
if(f)inc(i,0,n-1)a[i]=a[i]*invn%inf;
}



int main(){
n=read();
p[0]=p[1]=invp[0]=invp[1]=inv[1]=1;
inc(i,2,n)p[i]=p[i-1]*i%inf,inv[i]=inv[inf%i]*(inf-inf/i)%inf,invp[i]=invp[i-1]*inv[i]%inf;
inc(i,0,n)if(i&1)a[i]=inf-invp[i];else a[i]=invp[i];
b[0]=1;b[1]=n+1;inc(i,2,n)b[i]=(qpow(i,n+1)-1+inf)%inf*inv[i-1]%inf*invp[i]%inf;
init(n+n+1);
fft(a);fft(b);
inc(i,0,lim-1)a[i]=a[i]*b[i]%inf;
fft(a,1);
inc(i,0,n)ans+=qpow(2,i)*p[i]%inf*a[i]%inf,ans%=inf;
printf("%lld\n",ans);
return 0;
}