伯努利數的指數型生成函數為$B(x)={\sum }_{i=0}\frac{B_i}{i!}{x}^i=\frac{x}{e^x?1}$

由此可得$B_0=1,B_1=?\frac{1}{2},B_3=\frac{1}{6},B_4=0,B_5=\frac{1}{30}...$

如何求伯努利數？顯然可以多項式求逆，或者通過一個性質$O(n^2)$求。

（以下$[x^n]B(x)$指的是$B(x)$的$n$次項系數。）

因為$\frac{x}{e^x?1}(e^x?1)=x$

所以$[x^n]B(x)(e^x?1)={\sum }_{i=0}^{n?1}\frac{B_i}{i!}?\frac{1}{(n?i)!}=[n=1]$

所以${\sum }_{i=0}^{n?1}{B}_i{C}_n^i=[n=1]n!=[n=1]$

那么已知$B_0=1$，${\sum }_{i=0}^n{B}_i{C}_{n+1}^i=0$，則$B_n=?\frac{{\sum }_{i=0}^{n?1}{B}_i{C}_{n+1}^i}{n+1}$

于是就可以$O(n^2)$方便地求伯努利數，或者$O(n\log n)$常數巨大又不方便地求。

伯努利數還有個重要性質。

記$S_k(n)={\sum }_{i=0}^{n?1}{i}^k$

則$S_k(n)=\frac{1}{k+1}{\sum }_{i=0}^k{C}_{k+1}^i{B}_i{n}^{k+1?i}$

模板題1 51nod 1228 $O(n^2)$求伯努利數

#include<bits/stdc++.h> using namespace std; #define RI register int typedef long long LL; const int mod=1000000007; int B[2005],C[2005][2005],mi[2005],K,T;LL n;int qm(int x) {return x>=mod?x-mod:x;} int ksm(int x,int y) {int re=1;for(;y;y>>=1,x=1LL*x*x%mod) if(y&1) re=1LL*re*x%mod;return re; } void prework() {for(RI i=0;i<=2001;++i) {C[i][0]=1;for(RI j=1;j<=i;++j) C[i][j]=qm(C[i-1][j]+C[i-1][j-1]);}B[0]=1;for(RI i=1;i<=2000;++i) {for(RI j=0;j<i;++j) B[i]=qm(B[i]+1LL*B[j]*C[i+1][j]%mod);B[i]=qm(mod-1LL*B[i]*ksm(i+1,mod-2)%mod);} } int main() {prework();scanf("%d",&T);while(T--) {scanf("%lld%d",&n,&K);n=qm(n%mod+1);int ans=0;mi[0]=1;for(RI i=1;i<=K+1;++i) mi[i]=1LL*mi[i-1]*n%mod;for(RI i=0;i<=K;++i) ans=qm(ans+1LL*C[K+1][i]*B[i]%mod*mi[K+1-i]%mod);printf("%lld\n",1LL*ans*ksm(K+1,mod-2)%mod);}return 0; }

模板題2 51nod 1258 $O(n\log n)$求伯努利數

#include<bits/stdc++.h> using namespace std; #define RI register int typedef long long LL; typedef long double db; const int mod=1000000007,N=131080,M=32767; const db pi=acos(-1); int K,T;LL n; int fac[N],inv[N],ifac[N],rev[N],len[N]; int aa[N],bb[N],k1[N],k2[N]; struct com{db r,i;}a[N],b[N],Aa[N],Ab[N],Ba[N],Bb[N]; com operator + (com A,com B) {return (com){A.r+B.r,A.i+B.i};} com operator - (com A,com B) {return (com){A.r-B.r,A.i-B.i};} com operator * (com A,com B) {return (com){A.r*B.r-A.i*B.i,A.r*B.i+A.i*B.r};} com conj(com A) {return (com){A.r,-A.i};}int qm(int x) {return x>=mod?x-mod:x;} int ksm(int x,int y) {int re=1;for(;y;y>>=1,x=1LL*x*x%mod) if(y&1) re=1LL*re*x%mod;return re; } void FFT(com *a,int n) {for(RI i=0;i<n;++i) if(rev[i]>i) swap(a[i],a[rev[i]]);for(RI i=1;i<n;i<<=1) {com wn=(com){cos(pi/i),sin(pi/i)};for(RI j=0;j<n;j+=(i<<1)) {com t1,t2,w=(com){1,0};for(RI k=0;k<i;++k,w=w*wn)t1=a[j+k],t2=a[j+i+k]*w,a[j+k]=t1+t2,a[j+i+k]=t1-t2;}} } void mul(int *ka,int *kb,int *kc,int n) {for(RI i=0;i<n;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(len[n]-1));for(RI i=0;i<(n>>1);++i) {a[i]=(com){(db)(ka[i]&M),(db)(ka[i]>>15)};b[i]=(com){(db)(kb[i]&M),(db)(kb[i]>>15)};a[i+(n>>1)]=b[i+(n>>1)]=(com){0,0};}FFT(a,n),FFT(b,n);for(RI i=0;i<n;++i) {int j=(n-i)&(n-1);com kAa=(a[i]+conj(a[j]))*(com){0.5,0};com kAb=(a[i]-conj(a[j]))*(com){0,-0.5};com kBa=(b[i]+conj(b[j]))*(com){0.5,0};com kBb=(b[i]-conj(b[j]))*(com){0,-0.5};Aa[j]=kAa*kBa,Ab[j]=kAa*kBb,Ba[j]=kAb*kBa,Bb[j]=kAb*kBb;}for(RI i=0;i<n;++i)a[i]=Aa[i]+Ab[i]*(com){0,1},b[i]=Ba[i]+Bb[i]*(com){0,1};FFT(a,n),FFT(b,n);for(RI i=0;i<n;++i) {int kAa=(LL)(a[i].r/n+0.5)%mod;int kAb=(LL)(a[i].i/n+0.5)%mod;int kBa=(LL)(b[i].r/n+0.5)%mod;int kBb=(LL)(b[i].i/n+0.5)%mod;kc[i]=qm(((LL)kAa+((LL)(kAb+kBa)<<15)+((LL)kBb<<30))%mod+mod);} } void getinv(int *a,int *b,int n) {if(n==1) {b[0]=ksm(a[0],mod-2),b[1]=0;return;}getinv(a,b,n>>1),mul(a,b,k1,n<<1),mul(k1,b,k2,n<<1);for(RI i=0;i<n;++i) b[i]=qm(qm(b[i]+b[i])-k2[i]+mod),b[i+n]=0; }int C(int d,int u) {return 1LL*fac[d]*ifac[u]%mod*ifac[d-u]%mod;} void prework() {inv[0]=inv[1]=1,ifac[0]=1,fac[0]=1;for(RI i=1;i<N;++i) fac[i]=1LL*fac[i-1]*i%mod;for(RI i=2;i<N;++i) inv[i]=1LL*(mod-mod/i)*inv[mod%i]%mod;for(RI i=1;i<N;++i) ifac[i]=1LL*ifac[i-1]*inv[i]%mod;int kn=1;while(kn<=50000) kn<<=1,len[kn]=len[kn>>1]+1;len[kn<<1]=len[kn]+1;for(RI i=0;i<kn;++i) aa[i]=ifac[i+1];getinv(aa,bb,kn);for(RI i=0;i<=50000;++i) bb[i]=1LL*bb[i]*fac[i]%mod; } int main() {prework();scanf("%d",&T);while(T--) {scanf("%lld%d",&n,&K);n=qm(n%mod+1);int ans=0;for(RI i=1,j=n;i<=K+1;++i,j=1LL*j*n%mod)ans=qm(ans+1LL*C(K+1,K+1-i)*bb[K+1-i]%mod*j%mod);printf("%lld\n",1LL*ans*inv[K+1]%mod);}return 0; }

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