伯努利数

伯努利数

伯努利数，第(i)项记为(B_i)，是专门解决自然数幂求和而构造的一个数列
我们先记(S_k(n) = sumlimits_{i = 0}^{n - 1}i^k)
那么，不知道为什么

[S_k(n) = frac{1}{k + 1}sumlimits_{i = 0}^{k}{k + 1 choose i}B_in^{k + 1 - i} ]

我们发现(n = 0)是个特例
得到

[sumlimits_{i = 0}^{k}{k + 1 choose i}B_i = [k = 0] ]

所以我们有

[B_k = -frac{1}{k + 1} sumlimits_{i = 0}^{k - 1} {k + 1 choose i}B_i ]

就可以(O(k^2))递推辣
然后就可以(O(k))计算(S_k(n))辣

生成函数

伯努利数的指数型生成函数为

[egin{aligned} B(x) &= sumlimits_{i ge 0} B_{i} frac{x^{i}}{i!} \ &= frac{x}{e^{x} - 1} \ &= (sumlimits_{i ge 0}frac{x^{i}}{(i + 1)!})^{-1} end{aligned} ]

所以使用多项式求逆可以做到(O(klogk))预处理伯努利数

O(k^2)递推模板题

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2010,maxm = 100005,INF = 0x3f3f3f3f,P = 1000000007;
inline LL read(){
	LL out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
	return flag ? out : -out;
}
int fac[maxn],fv[maxn],inv[maxn],B[maxn],K,N = 2005;
LL n;
inline int qpow(int a,int b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
inline int C(int n,int m){
	if (m > n) return 0;
	return 1ll * fac[n] * fv[m] % P * fv[n - m] % P;
}
void init(){
	fac[0] = fac[1] = fv[0] = fv[1] = inv[0] = inv[1] = 1;
	for (int i = 2; i <= N; i++){
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
		fv[i] = 1ll * fv[i - 1] * inv[i] % P;
	}
	B[0] = 1;
	for (int k = 1; k < N; k++){
		for (int i = 0; i < k; i++)
			B[k] = (B[k] + 1ll * C(k + 1,i) * B[i] % P) % P;
		B[k] = 1ll * B[k] * (P - 1) % P * inv[k + 1] % P;
	}
}
int main(){
	init();
	int T = read();
	while (T--){
		n = read() % P; K = read();
		int tmp = n,ans = 0;
		for (int i = K; ~i; i--){
			ans = (ans + 1ll * C(K + 1,i) * B[i] % P * tmp % P) % P;
			tmp = 1ll * tmp * n % P;
		}
		ans = 1ll * ans * inv[K + 1] % P;
		printf("%d
",(ans + qpow(n,K)) % P);
	}
	return 0;
}