51nod1228 序列求和（伯努利数）

题面

传送门

题解

\(O(n^2)\)预处理伯努利数

不知道伯努利数是什么的可以看看这篇文章

不过这个数据范围拉格朗日差值应该也没问题……吧……大概……

//minamoto

#include<bits/stdc++.h>

#define R register

#define ll long long

#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)

#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)

#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)

using namespace std;

char buf[1<<21],*p1=buf,*p2=buf;

inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}

ll read(){

    R ll res,f=1;R char ch;

    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);

    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');

    return res*f;

}

char sr[1<<21],z[20];int K=-1,Z=0;

inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}

void print(R int x){

    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;

    while(z[++Z]=x%10+48,x/=10);

    while(sr[++K]=z[Z],--Z);sr[++K]='\n';

}

const int N=2005,P=1e9+7;

inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}

inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}

inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}

int ksm(R int x,R int y){

	R int res=1;

	for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);

	return res;

}

int fac[N],ifac[N],inv[N],B[N];

inline int C(R int n,R int m){return m>n?0:1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}

ll n;int k,T,nw,q,res;

void init(int n){

	B[0]=fac[0]=fac[1]=ifac[0]=ifac[1]=inv[0]=inv[1]=1;

	fp(i,2,n){

		fac[i]=mul(fac[i-1],i),

		inv[i]=1ll*(P-P/i)*inv[P%i]%P,

		ifac[i]=mul(ifac[i-1],inv[i]);

	}

	fp(i,1,n){

		fp(j,0,i-1)B[i]=dec(B[i],mul(B[j],C(i+1,j)));

		B[i]=mul(B[i],inv[i+1]);

	}

}

int main(){

//	freopen("testdata.in","r",stdin);

	init(N-1);

	for(int T=read();T;--T){

		n=read(),k=read(),nw=q=(n+1)%P,res=0;

		for(R int i=k;~i;--i,nw=mul(nw,q))res=add(res,1ll*B[i]*C(k+1,i)%P*nw%P);

		res=mul(res,inv[k+1]);

		print(res);

	}

	return Ot(),0;

}