【51Nod1258】序列求和V4（FFT）

【51Nod1258】序列求和V4（FFT）

题面

51Nod

多组数据，求：

\[Ans=\sum_{i=1}^ni^k,n\le 10^{18},k\le50000
\]

题解

预处理伯努利数，时间复杂度\(O(nlogn)\)

然后利用伯努利数求和即可。

\[\sum_{i=1}^n i^k=\frac{1}{k+1}\sum_{i=0}^kB_iC_{k+1}^i(n+1)^{k+1-i}
\]

预处理需要多项式求逆，因为模数不太好，所以需要\(MTT\)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MOD 1000000007
#define MAX 150000
const int NN=50000;
const int M=sqrt(MOD);
const double Pi=acos(-1);
inline ll read()
{
    RG ll x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}
int fpow(int a,int b){int s=1;while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}return s;}
struct Complex{double a,b;}W[MAX],A1[MAX],A2[MAX],B1[MAX],B2[MAX],X[MAX],Y[MAX],Z[MAX];
Complex operator+(Complex a,Complex b){return (Complex){a.a+b.a,a.b+b.b};}
Complex operator-(Complex a,Complex b){return (Complex){a.a-b.a,a.b-b.b};}
Complex operator*(Complex a,Complex b){return (Complex){a.a*b.a-a.b*b.b,a.b*b.a+a.a*b.b};}
int r[MAX],N,l;
void FFT(Complex *P,int N,int opt)
{
	for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
	for(int i=1;i<N;i<<=1)
		for(int p=i<<1,j=0;j<N;j+=p)
			for(int k=0;k<i;++k)
			{
				Complex w=(Complex){W[N/i*k].a,W[N/i*k].b*opt};
				Complex X=P[j+k],Y=P[i+j+k]*w;
				P[j+k]=X+Y;P[i+j+k]=X-Y;
			}
	if(opt==-1)for(int i=0;i<N;++i)P[i].a/=N;
}
void MTT(int *a,int *b,int len,int *c)
{
	memset(A1,0,sizeof(A1));memset(B1,0,sizeof(B1));
	memset(A2,0,sizeof(A2));memset(B2,0,sizeof(B2));
	for(N=1,l=0;N<=len;N<<=1)++l;
	for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	for(int i=1;i<N;i<<=1)
		for(int k=0;k<i;++k)W[N/i*k]=(Complex){cos(k*Pi/i),sin(k*Pi/i)};
	for(int i=0;i<len;++i)a[i]%=MOD,b[i]%=MOD;
	for(int i=0;i<len;++i)A1[i].a=a[i]/M,A2[i].a=a[i]%M;
	for(int i=0;i<len;++i)B1[i].a=b[i]/M,B2[i].a=b[i]%M;
	memset(X,0,sizeof(X));memset(Y,0,sizeof(Y));memset(Z,0,sizeof(Z));
	FFT(A1,N,1);FFT(A2,N,1);FFT(B1,N,1);FFT(B2,N,1);
	for(int i=0;i<N;++i)
	{
		X[i]=A1[i]*B1[i];
		Y[i]=A1[i]*B2[i]+A2[i]*B1[i];
		Z[i]=A2[i]*B2[i];
	}
	FFT(X,N,-1);FFT(Y,N,-1);FFT(Z,N,-1);
	for(int i=0;i<len;++i)
	{
		int ans=0;
		ans=(ll)(X[i].a+0.5)%MOD*M%MOD*M%MOD;
		ans=(ans+(ll)(Y[i].a+0.5)%MOD*M)%MOD;
		ans=(ans+(ll)(Z[i].a+0.5))%MOD;
		c[i]=ans;
	}
}
int c[MAX],d[MAX];
void Inv(int *a,int *b,int len)
{
	if(len==1){b[0]=fpow(a[0],MOD-2);return;}
	Inv(a,b,len>>1);
	MTT(a,b,len,c);MTT(b,c,len,d);
	for(int i=0;i<len;++i)b[i]=(b[i]+b[i])%MOD;
	for(int i=0;i<len;++i)b[i]=(b[i]+MOD-d[i])%MOD;
}
int a[MAX];
int n,B[MAX],jc[MAX],jv[MAX],inv[MAX];
int C(int n,int m){return 1ll*jc[n]*jv[m]%MOD*jv[n-m]%MOD;}
int main()
{
	B[0]=jc[0]=jv[0]=inv[0]=inv[1]=1;
	for(int i=2;i<=NN+NN;++i)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=1;i<=NN+NN;++i)jc[i]=1ll*jc[i-1]*i%MOD;
	for(int i=1;i<=NN+NN;++i)jv[i]=1ll*jv[i-1]*inv[i]%MOD;
	for(int i=0;i<=NN;++i)a[i]=jv[i+1];
	Inv(a,B,1<<16);
	for(int i=0;i<=NN;++i)B[i]=1ll*B[i]*jc[i]%MOD;
	int T=read();
	while(T--)
	{
		int n=read()%MOD,k=read(),ans=0,nw=n+1;
		for(int i=k;~i;--i,nw=1ll*nw*(n+1)%MOD)ans=(ans+1ll*C(k+1,i)*B[i]%MOD*nw)%MOD;
		ans=1ll*ans*inv[k+1]%MOD;
		printf("%d\n",ans);
	}
	return 0;
}