伯努利数的应用

Posted 2020-10-24 Stump

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;
}