伯努利数
伯努利数
\(B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=\frac{1}{30}\)
可以利用下面的式子计算。
\[B_0=1,\sum_{i=0}^nB_iC_{n+1}^i=0
\]
转化：
\[\begin{aligned}
\sum_{i=0}^nB_iC_{n+1}^i&=0\\
\sum_{i=0}^{n-1}B_iC_{n}^i&=0\\
\sum_{i=0}^{n-1}B_iC_{n}^i+B_n&=B_n(n>1)\\
\sum_{i=0}^{n}B_iC_n^i&=B_n(n>1)\\
\sum_{i=0}^n\frac{n!}{i!(n-i)!}B_i&=B_n(n>1)\\
\sum_{i=0}^n\frac{1}{i!(n-i)!}B_i&=\frac{B_n}{n!}(n>1)\\
\end{aligned}
\]
对于\(B_i\)构建指数型生成函数$$B(x)=\sum_{i=0}{\infty}\frac{B_i}{i!}xi$$
发现上面转化出来的左侧是\(B(x)\)乘上\(\{1\}\)的指数型生成函数。
所以我们可以得到式子$$B(x)e^x=B(x)+x$$
需要额外加\(x\)的原因是上面转化式子中的右侧没有\(B_1\)
所以我们可以得到伯努利数的另外一种形式\(B(x)=\frac{x}{e^x-1}\)
所以我们利用多项式求逆在\(O(nlogn)\)计算伯努利数。
当然，这个式子也可以拆开，然后\(O(n^2)\)计算。
\[\begin{aligned}
\sum_{i=0}^nB_iC_{n+1}^i&=0\\
\sum_{i=0}^{n-1}B_iC_{n+1}^i+B_nC_{n+1}^n&=0\\
-B_n(n+1)&=\sum_{i=0}^{n-1}B_iC_{n+1}^i\\
B_n&=-\frac{1}{n+1}\sum_{i=0}^{n-1}B_iC_{n+1}^i
\end{aligned}
\]
伯努利数可以用来计算自然数的幂和。
\[\begin{aligned}
S_k(n)&=\sum_{i=0}^{n-1}i^k\\
&=\frac{1}{k+1}\sum_{i=0}^kC_{k+1}^iB_in^{k+1-i}
\end{aligned}
\]
这个式子是这么来的
\[\begin{aligned}
A(x)&=\sum_{i=0}^{\infty}(\sum_{j=0}^{n-1}j^i)\frac{x^i}{i!}=\sum_{j=0}^{n-1}\sum_{i=0}^{\infty}j^i\frac{x^i}{i!}\\
&=\sum_{j=0}^{n-1}e^{jx}=\frac{e^{nx}-1}{e^x-1}\\
&=B(x)\frac{e^{nx}-1}{x}
\end{aligned}
\]
然后再拆掉这个式子就可以得到上面的结果（我说其实我并不会拆这个式子）
以下是\(51Nod1228\)，\(O(n^2)\)预处理伯努利数
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MOD 1000000007
#define MAX 2500
inline ll read()
{
RG ll x=0,t=1;RG char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-')t=-1,ch=getchar();
while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
return x*t;
}
int fpow(int a,int b)
{
int s=1;
while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}
return s;
}
int B[MAX],jc[MAX],jv[MAX],inv[MAX];
int C(int n,int m){return 1ll*jc[n]*jv[m]%MOD*jv[n-m]%MOD;}
int main()
{
B[0]=jc[0]=jv[0]=inv[0]=inv[1]=1;
for(int i=2;i<MAX;++i)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
for(int i=1;i<MAX;++i)jc[i]=1ll*jc[i-1]*i%MOD;
for(int i=1;i<MAX;++i)jv[i]=1ll*jv[i-1]*inv[i]%MOD;
for(int i=1;i<MAX;++i)
{
for(int j=0;j<i;++j)B[i]=(B[i]+1ll*B[j]*C(i+1,j))%MOD;
B[i]=1ll*B[i]*inv[i+1]%MOD;B[i]=(MOD-B[i])%MOD;
}
int T=read();
while(T--)
{
ll n=read();int k=read(),ans=0;
int nw=(n+1)%MOD,q=(n+1)%MOD;
for(int i=k;~i;--i,nw=1ll*nw*q%MOD)ans=(ans+1ll*C(k+1,i)*B[i]%MOD*nw)%MOD;
ans=1ll*ans*inv[k+1]%MOD;
printf("%d\n",ans);
}
return 0;
}
一下是\(51Nod1258\)，多项式求逆预处理伯努利数
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MOD 1000000007
#define MAX 150000
const int NN=50000;
const int M=sqrt(MOD);
const double Pi=acos(-1);
inline ll read()
{
RG ll x=0,t=1;RG char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-')t=-1,ch=getchar();
while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
return x*t;
}
int fpow(int a,int b){int s=1;while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}return s;}
struct Complex{double a,b;}W[MAX],A1[MAX],A2[MAX],B1[MAX],B2[MAX],X[MAX],Y[MAX],Z[MAX];
Complex operator+(Complex a,Complex b){return (Complex){a.a+b.a,a.b+b.b};}
Complex operator-(Complex a,Complex b){return (Complex){a.a-b.a,a.b-b.b};}
Complex operator*(Complex a,Complex b){return (Complex){a.a*b.a-a.b*b.b,a.b*b.a+a.a*b.b};}
int r[MAX],N,l;
void FFT(Complex *P,int N,int opt)
{
for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
for(int i=1;i<N;i<<=1)
for(int p=i<<1,j=0;j<N;j+=p)
for(int k=0;k<i;++k)
{
Complex w=(Complex){W[N/i*k].a,W[N/i*k].b*opt};
Complex X=P[j+k],Y=P[i+j+k]*w;
P[j+k]=X+Y;P[i+j+k]=X-Y;
}
if(opt==-1)for(int i=0;i<N;++i)P[i].a/=N;
}
void MTT(int *a,int *b,int len,int *c)
{
memset(A1,0,sizeof(A1));memset(B1,0,sizeof(B1));
memset(A2,0,sizeof(A2));memset(B2,0,sizeof(B2));
for(N=1,l=0;N<=len;N<<=1)++l;
for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
for(int i=1;i<N;i<<=1)
for(int k=0;k<i;++k)W[N/i*k]=(Complex){cos(k*Pi/i),sin(k*Pi/i)};
for(int i=0;i<len;++i)a[i]%=MOD,b[i]%=MOD;
for(int i=0;i<len;++i)A1[i].a=a[i]/M,A2[i].a=a[i]%M;
for(int i=0;i<len;++i)B1[i].a=b[i]/M,B2[i].a=b[i]%M;
memset(X,0,sizeof(X));memset(Y,0,sizeof(Y));memset(Z,0,sizeof(Z));
FFT(A1,N,1);FFT(A2,N,1);FFT(B1,N,1);FFT(B2,N,1);
for(int i=0;i<N;++i)
{
X[i]=A1[i]*B1[i];
Y[i]=A1[i]*B2[i]+A2[i]*B1[i];
Z[i]=A2[i]*B2[i];
}
FFT(X,N,-1);FFT(Y,N,-1);FFT(Z,N,-1);
for(int i=0;i<len;++i)
{
int ans=0;
ans=(ll)(X[i].a+0.5)%MOD*M%MOD*M%MOD;
ans=(ans+(ll)(Y[i].a+0.5)%MOD*M)%MOD;
ans=(ans+(ll)(Z[i].a+0.5))%MOD;
c[i]=ans;
}
}
int c[MAX],d[MAX];
void Inv(int *a,int *b,int len)
{
if(len==1){b[0]=fpow(a[0],MOD-2);return;}
Inv(a,b,len>>1);
MTT(a,b,len,c);MTT(b,c,len,d);
for(int i=0;i<len;++i)b[i]=(b[i]+b[i])%MOD;
for(int i=0;i<len;++i)b[i]=(b[i]+MOD-d[i])%MOD;
}
int a[MAX];
int n,B[MAX],jc[MAX],jv[MAX],inv[MAX];
int C(int n,int m){return 1ll*jc[n]*jv[m]%MOD*jv[n-m]%MOD;}
int main()
{
B[0]=jc[0]=jv[0]=inv[0]=inv[1]=1;
for(int i=2;i<=NN+NN;++i)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
for(int i=1;i<=NN+NN;++i)jc[i]=1ll*jc[i-1]*i%MOD;
for(int i=1;i<=NN+NN;++i)jv[i]=1ll*jv[i-1]*inv[i]%MOD;
for(int i=0;i<=NN;++i)a[i]=jv[i+1];
Inv(a,B,1<<16);
for(int i=0;i<=NN;++i)B[i]=1ll*B[i]*jc[i]%MOD;
int T=read();
while(T--)
{
int n=read()%MOD,k=read(),ans=0,nw=n+1;
for(int i=k;~i;--i,nw=1ll*nw*(n+1)%MOD)ans=(ans+1ll*C(k+1,i)*B[i]%MOD*nw)%MOD;
ans=1ll*ans*inv[k+1]%MOD;
printf("%d\n",ans);
}
return 0;
}
