51nod1228
http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1228
#include<cstdio>
typedef long long ll;
const int maxn=5005,mod=1e9+7;
int c[maxn][maxn],b[maxn],inv[maxn];
int T,k,tmp,ans;
ll n;
int main(){
for(register int i=0;i<=5000;++i){
c[i][0]=1;
for(register int j=1;j<=i;++j)
c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
}
inv[1]=1;
for(register int i=2;i<=5000;++i)
inv[i]=mod-1ll*mod/i*inv[mod%i]%mod;
b[0]=1;
for(register int i=1;i<=5000;++i){
for(register int j=0;j<i;++j)
b[i]=(b[i]+1ll*c[i+1][j]*b[j]%mod)%mod;
b[i]=(mod-1ll*b[i]*inv[i+1]%mod)%mod;
}
scanf("%d",&T);
while(T--){
scanf("%lld%d",&n,&k);
++n;
n%=mod;
tmp=n;
ans=0;
for(register int i=1;i<=k+1;++i,tmp=1ll*tmp*n%mod)
ans=(ans+1ll*c[k+1][i]*b[k+1-i]%mod*tmp%mod)%mod;
ans=1ll*ans*inv[k+1]%mod;
printf("%d\n",ans);
}
return 0;
}
fft做多项式求逆，求伯努利数
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod=998244353,maxn=2e5+5;
int a[maxn],b[maxn],tmp[maxn],s[maxn],gn[maxn],inv[maxn],f[maxn];
int n,k;
inline int fp(int a,int b){
int ret=1;
while(b){
if(b&1)ret=1ll*a*ret%mod;
a=1ll*a*a%mod;b>>=1;
}
return ret;
}
inline void ntt(int *a,int p,int f){
for(register int i=0;i<p;++i)
if(i<s[i])
swap(a[i],a[s[i]]);
for(register int i=1,t=0,g,w,v;i<p;i<<=1,++t){
g=gn[t];
for(register int j=0;j<p;j+=(i<<1)){
w=1;
for(register int k=j;k<i+j;++k,w=1ll*w*g%mod){
v=1ll*w*a[i+k]%mod;
a[i+k]=(a[k]-v+mod)%mod;
a[k]=(a[k]+v)%mod;
}
}
}
if(f==1)return;
reverse(a+1,a+p);
int ny=fp(p,mod-2);
for(register int i=0;i<p;++i)
a[i]=1ll*a[i]*ny%mod;
}
inline void solve(int *b,int deg){
if(deg==1){
b[0]=fp(a[0],mod-2);
return;
}
solve(b,(deg+1)>>1);
int p=1,lg2=0;while(p<(deg<<1))p<<=1,++lg2;
for(register int i=0;i<p;++i)tmp[i]=i<deg?a[i]:0;
for(register int i=((deg+1)>>1);i<p;++i)b[i]=0;
for(register int i=0;i<p;++i)s[i]=(s[i>>1]>>1)^((i&1)<<(lg2-1));
ntt(tmp,p,1),ntt(b,p,1);
for(register int i=0;i<p;++i)b[i]=(2ll*b[i]%mod-1ll*tmp[i]*b[i]%mod*b[i]%mod+mod)%mod;
ntt(b,p,-1);
}
int main(){
for(register int t=0,i=1;t<=20;i<<=1,++t)
gn[t]=fp(3,(mod-1)/(i<<1));
scanf("%d",&n);inv[1]=1;a[0]=1;f[0]=1;
for(register int i=2;i<=n+1;++i)inv[i]=mod-1ll*mod/i*inv[mod%i]%mod;
for(register int i=1;i<=n;++i)a[i]=1ll*a[i-1]*inv[i+1]%mod;
solve(b,n+1);
for(register int i=1;i<=n;++i)f[i]=1ll*f[i-1]*i%mod,b[i]=1ll*b[i]*f[i]%mod;
for(register int i=0;i<=n;++i)printf("%d ",b[i]);
return 0;
}
51nod1258
http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1258
加强版，推荐写插值法，别写NTT+CRT
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e6+11,mod=1e9+7;
int f[maxn],fac[maxn],p[maxn],q[maxn],inv[maxn];
int T,k,ans;
ll n;
inline int fp(int a,int b){
int ret=1;
while(b){
if(b&1)ret=1ll*ret*a%mod;
a=1ll*a*a%mod;b>>=1;
}
return ret;
}
#define gc getchar()
inline int read(){
char c;while(c=gc,c==' '||c=='\n');int data=c-48;
while(c=gc,c>='0'&&c<='9')data=(c-48+(data<<1)%mod+((1ll*data)<<3)%mod)%mod;;return data;
}
int main(){
fac[0]=1;
for(register int i=1;i<=60000;++i)
fac[i]=1ll*fac[i-1]*i%mod;
inv[1]=1;for(register int i=2;i<=60000;++i)inv[i]=mod-1ll*mod/i*inv[mod%i]%mod;
inv[0]=1;for(register int i=2;i<=60000;++i)inv[i]=1ll*inv[i]*inv[i-1]%mod;
T=read();
while(T--){
n=read();k=read();ans=0;
for(register int i=1;i<=k+2;++i)f[i]=(f[i-1]+fp(i,k))%mod;
p[0]=1;for(register int i=1;i<=k+2;++i)p[i]=1ll*p[i-1]*(n-i)%mod;
q[k+3]=1;for(register int i=k+2;i;--i)q[i]=1ll*q[i+1]*(n-i)%mod;
for(register int i=1;i<=k+2;++i)ans=(ans+((k-i)&1?(-1ll):1ll)*f[i]*p[i-1]%mod*q[i+1]%mod*inv[i-1]%mod*inv[k+2-i]%mod+mod)%mod;
printf("%d\n",ans);
}
return 0;
} 
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e6+11,mod=1e9+7;
int f[maxn],fac[maxn],p[maxn],q[maxn],inv[maxn],pr[maxn],fr[maxn];
bool np[maxn];
int T,k,ans;
ll n;
inline int fp(int a,int b){
int ret=1;
while(b){
if(b&1)ret=1ll*ret*a%mod;
a=1ll*a*a%mod;b>>=1;
}
return ret;
}
inline void shai_fa(){
f[1]=1;
for(register int i=2;i<=60000;++i){
if(!np[i]){
pr[++pr[0]]=i;
fr[i]=i;
}
for(register int j=1;j<=pr[0]&&1ll*pr[j]*i<=60000;++j){
np[i*pr[j]]=1;
fr[i*pr[j]]=pr[j];
if(i%pr[j]==0)break;
}
}
}
int main(){
fac[0]=1;for(register int i=1;i<=60000;++i)fac[i]=1ll*fac[i-1]*i%mod;
inv[1]=1;for(register int i=2;i<=60000;++i)inv[i]=mod-1ll*mod/i*inv[mod%i]%mod;
inv[0]=1;for(register int i=2;i<=60000;++i)inv[i]=1ll*inv[i]*inv[i-1]%mod;
scanf("%d",&T);
shai_fa();
while(T--){
scanf("%lld%d",&n,&k);ans=0;n%=mod;
for(register int i=2;i<=k+2;++i)f[i]=(fr[i]==i)?fp(i,k):1ll*f[i/fr[i]]*f[fr[i]]%mod;
for(register int i=2;i<=k+2;++i)f[i]=(f[i]+f[i-1])%mod;
p[0]=1;for(register int i=1;i<=k+2;++i)p[i]=1ll*p[i-1]*(n-i+mod)%mod;
q[k+3]=1;for(register int i=k+2;i;--i)q[i]=1ll*q[i+1]*(n-i+mod)%mod;
for(register int i=1;i<=k+2;++i)ans=(ans+((k-i)&1?(-1ll):1ll)*f[i]*p[i-1]%mod*q[i+1]%mod*inv[i-1]%mod*inv[k+2-i]%mod+mod)%mod;
printf("%d\n",ans);
}
return 0;
}
来源：oschina
链接：https://my.oschina.net/u/4393219/blog/4237222