题面

题解

不知道伯努利数是什么的可以先去看看

多项式求逆预处理伯努利数就行

因为这里模数感人，所以得用\(MTT\)

//minamoto#include
    
     #define R register#define ll long long#define fp(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;}const int N=5e5+5,K=50005,P=1e9+7;const double Pi=acos(-1.0);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;}struct cp{    double x,y;    cp(){}    cp(R double xx,R double yy):x(xx),y(yy){}    inline cp operator +(const cp &b)const{return cp(x+b.x,y+b.y);}    inline cp operator -(const cp &b)const{return cp(x-b.x,y-b.y);}    inline cp operator *(const cp &b)const{return cp(x*b.x-y*b.y,x*b.y+y*b.x);}}w[2][N];int r[N],inv[N],ifac[N],fac[N],B[N],A[N],c[N],d[N],lim,l,n,k,res,nw;inline void init(int len){    lim=1,l=0;while(lim
      
       <<=1,++l;    fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));}void FFT(cp *A,int ty){    fp(i,0,lim-1)if(i
       
        >15,B[i].x=a[i]&32767,        C[i].x=b[i]>>15,D[i].x=b[i]&32767,        A[i].y=B[i].y=C[i].y=D[i].y=0;    }fp(i,len,lim-1)A[i]=B[i]=C[i]=D[i]=cp(0,0);    FFT(A,1),FFT(B,1),FFT(C,1),FFT(D,1);    fp(i,0,lim-1)E[i]=A[i]*C[i],F[i]=A[i]*D[i]+B[i]*C[i],G[i]=B[i]*D[i];    FFT(E,0),FFT(F,0),FFT(G,0);    fp(i,0,lim-1)c[i]=(((ll)(E[i].x+0.5)%P<<30)+((ll)(F[i].x+0.5)<<15)+((ll)(G[i].x+0.5)))%P;}void Inv(int *a,int *b,int len){    if(len==1)return b[0]=ksm(a[0],P-2),void();    Inv(a,b,len>>1);    MTT(a,b,len,c),MTT(c,b,len,d);    fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),d[i]);}inline void init(){    for(R int i=1;i<(1<<17);i<<=1)fp(k,0,i-1)        w[1][i+k]=cp(cos(Pi*k/i),sin(Pi*k/i)),w[0][i+k]=cp(cos(Pi*k/i),-sin(Pi*k/i));    B[0]=ifac[0]=ifac[1]=inv[0]=inv[1]=fac[0]=fac[1]=1;    fp(i,2,K+5){        fac[i]=mul(fac[i-1],i),        inv[i]=mul(P-P/i,inv[P%i]),        ifac[i]=mul(ifac[i-1],inv[i]);    }    fp(i,0,K)A[i]=ifac[i+1];    Inv(A,B,1<<16);    fp(i,0,K)B[i]=mul(B[i],fac[i]);}inline int C(R int n,R int m){return m>n?0:1ll*fac[n]*ifac[m]%P*ifac[n-m]%P;}int main(){//  freopen("testdata.in","r",stdin);    init();    for(int T=read();T;--T){        n=read()%P,k=read(),res=0,nw=n+1;        for(R int i=k;~i;--i,nw=mul(nw,n+1))res=add(res,1ll*C(k+1,i)*B[i]%P*nw%P);        res=mul(res,inv[k+1]),printf("%d\n",res);    }    return 0;}