# 伯努利数学习笔记&&Luogu P3711 仓鼠的数学题

Luogu P3711

### 伯努利数

$$\sum_{i=0}^n B_i\binom{n+1}{i}=0$$

$$S_{k,n-1}=\sum_{i=0}^{n-1}i^k=\frac{1}{k+1}\sum_{i=0}^k\binom{k+1}{i}B_in^{k+1-i}$$

\begin{aligned} \sum_{i=0}^n B_i\binom{n+1}{i}&=0\\ \sum_{i=0}^{n-1} B_i\binom{n}{i}&=0 \ (n>1)\\ B_n+\sum_{i=0}^{n-1} B_i\binom{n}{i}&=B_n \ (n>1)\\ B_n&=\sum_{i=0}^nB_i\binom{n}{i} \ (n>1)\\ \frac{B_n}{n!}&=\sum_{i=0}^n\frac{B_i}{i!(n-i)!}\\ \end{aligned}

### 回到原题

\begin{aligned} ans&=\sum_{k=0}^na_k S_{k,x}\\ &=\sum_{k=0}^na_k(x^k+\frac{1}{k+1}\sum_{i=0}^k\binom{k+1}{i}B_ix^{k+1-i})\\ &=(\sum_{k=0}^na_kx^k)+(\sum_{k=0}^nk!\sum_{i=0}^k\frac{B_i}{i!(k+1-i)!}x^{k+1-i})\\ ans[x^d]&=a_d+\sum_{i=0}^{n+1}\frac{B_i}{d!i!}(d+i-1)!\\ \frac{ans[x^d]}{d!}&=a_d+\sum_{i=0}^{n+1}\frac{B_i}{i!}(d+i-1)! \end{aligned}

### 代码

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#define p 998244353
#define rt register int
#define ll long long
#define ull unsigned long long
using namespace std;
ll x=0;char zf=1;char ch=getchar();
while(ch!='-'&&!isdigit(ch))ch=getchar();
if(ch=='-')zf=-1,ch=getchar();
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();return x*zf;
}
void write(ll y){if(y<0)putchar('-'),y=-y;if(y>9)write(y/10);putchar(y%10+48);}
void writeln(const ll y){write(y);putchar('\n');}
int k,m,n,x,y,z,cnt,ans;

namespace Poly{
#define poly vector<int>
#define MAXN 524288
int ksm(int x,int y=p-2){
int ans=1;
for(;y;y>>=1,x=1ll*x*x%p)if(y&1)ans=1ll*ans*x%p;
return ans;
}
void NTT(int n,poly &A,int fla){
static ull F[MAXN],W[MAXN];A.resize(n);
for(rt i=0,j=0;i<n;i++){
F[i]=A[j];
for(rt k=n>>1;(j^=k)<k;k>>=1);
}
for(rt i=1;i<n;i<<=1){
const int w=W[1]=ksm(3,(p-1)/2/i);W[0]=1;
for(rt k=2;k<i;k++)W[k]=1ll*W[k-1]*w%p;
for(rt j=0;j<n;j+=i<<1){
for(rt k=0;k<i;k++){
const ull x=F[j+k],y=F[i+j+k]*W[k]%p;
F[j+k]=x+y,F[i+j+k]=x+p-y;
}
}
}
for(rt i=0;i<n;i++)A[i]=F[i]%p;
if(fla==-1){
const int invn=ksm(n);
reverse(A.begin()+1,A.end());
for(rt i=0;i<n;i++)A[i]=1ll*A[i]*invn%p;
}
}
poly Mul(poly x,poly y){
int sz=x.size()+y.size()-1,lim=1;
while(lim<=sz)lim<<=1;
NTT(lim,x,1);NTT(lim,y,1);
for(rt i=0;i<lim;i++)x[i]=1ll*x[i]*y[i]%p;
NTT(lim,x,-1);x.resize(sz);return x;
}
poly Inv(poly a,int n=-1){
if(n==-1)n=a.size();
if(n==1)return {ksm(a[0])};
poly c=Inv(a,n+1>>1),d(&a[0],&a[n]);
int lim=1;while(lim<=n*2)lim<<=1;
NTT(lim,c,1);NTT(lim,d,1);
for(rt i=0;i<lim;i++)c[i]=1ll*c[i]*(2ll+p-1ll*d[i]*c[i]%p)%p;
NTT(lim,c,-1);c.resize(n);return c;
}
}
using namespace Poly;
int inv[250010],jc[250010],njc[250010],a[250010];
poly B;
void init(int k){
for(rt i=0;i<=1;i++)inv[i]=jc[i]=njc[i]=1;
for(rt i=2;i<=k+2;i++){
inv[i]=1ll*inv[p%i]*(p-p/i)%p;
jc[i]=1ll*jc[i-1]*i%p;
njc[i]=1ll*njc[i-1]*inv[i]%p;
}
B.resize(k+1);
for(rt i=0;i<=k;i++)B[i]=njc[i+1];
B=Inv(B);
for(rt i=0;i<=k;i++)B[i]=1ll*B[i]*jc[i]%p;
}
int main(){
poly ans(n+2),C(n+1);
for(rt i=0;i<=n;i++)B[i]=1ll*B[i]*njc[i]%p;
for(rt i=0;i<=n;i++)C[i]=1ll*jc[i]*a[i]%p;
reverse(&B[0],&B[n+1]);B.resize(n+1);C.resize(n+1);
ans=Mul(B,C);
for(rt i=0;i<=n+1;i++)ans[n+i-1]=1ll*ans[n+i-1]*njc[i]%p;
for(rt i=0;i<=n;i++)(ans[n+i-1]+=a[i])%=p;write(a[0]),putchar(' ');
for(rt i=1;i<=n+1;i++)write(ans[n+i-1]),putchar(' ');
return 0;
}

posted @ 2019-03-13 19:07  Kananix  阅读(254)  评论(0编辑  收藏