标签：仓鼠 frac int lim sum 1ll P3711 例题 mod
伯努利数
定义 S k ( n ) = ∑ i = 0 n − 1 i k S_k(n) = \sum\limits_{i = 0} ^{n - 1} i ^ k Sk(n)=i=0∑n−1ik。
从二项式出发
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+
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i
+
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1
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1
把
次
方
k
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的
移
项
，
再
整
体
相
加
，
得
n
k
+
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=
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k
C
k
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S
i
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n
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S
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+
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S
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S
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(0 + 1) ^ {k + 1} = \sum_{i = 0} ^{k} C_{k + 1} ^{i} 0 ^ {i} + 0 ^{k + 1}\\ \vdots\\ (n - 1 + 1) ^{k + 1} = \sum_{i = 0} ^{k} C_{k + 1} ^{i} (n - 1) ^{i} + (n - 1) ^{k + 1}\\ 把次方k + 1的移项，再整体相加，得\\ n ^{k + 1} = \sum_{i = 0} ^{k} C_{k + 1} ^{i} S_i(n)\\ n ^{k + 1} = \sum_{i = 0} ^{k - 1} C_{k + 1} ^{i} S_i(n) + (k + 1) S_k(n)\\ S_k(n) = \frac{1}{k + 1}\left(n ^{k + 1} - \sum_{i = 0} ^{k - 1} C_{k + 1} ^{i} S_i(n) \right)\\
(0+1)k+1=i=0∑kCk+1i0i+0k+1⋮(n−1+1)k+1=i=0∑kCk+1i(n−1)i+(n−1)k+1把次方k+1的移项，再整体相加，得nk+1=i=0∑kCk+1iSi(n)nk+1=i=0∑k−1Ck+1iSi(n)+(k+1)Sk(n)Sk(n)=k+11(nk+1−i=0∑k−1Ck+1iSi(n))
于是我们有了一个
O
(
n
2
)
O(n ^ 2)
O(n2)递推求
S
i
(
n
)
S_i(n)
Si(n)得算法，
对二项式再进一步考虑：
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两
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同
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相
加
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求
导
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，
代
入
S
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′
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m
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=
∑
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k
i
S
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0
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k
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从
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−
>
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求
积
分
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k
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−
S
k
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0
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=
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k
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=
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于
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≠
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+
1
,
所
以
0
k
−
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+
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S
k
(
n
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=
∑
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=
0
k
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k
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1
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k
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S
i
′
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0
)
n
k
+
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−
1
(0 + m) ^{k} = \sum_{i = 0} ^{k} C_{k} ^{i} 0 ^{i} m ^{k - i}\\ \vdots\\ (n - 1 + m) ^{k} = \sum_{i = 0} ^{k} C_{k} ^{i}(n - 1) ^{i} m ^{k - i}\\ 左右两边同时相加\\ S_k(n + m) - S_k(m) = \sum_{i = 0} ^{k} C_{k} ^{i} S_{i} (n) m ^{k - i}\\ 左右两边对n求导\\ S'_k(n + m) = \sum_{i = 0} ^{k} C_{k} ^{i} S'_i(n) m^{k - i}\\ \\ n = 0，代入\\ S'_k(m) = \sum_{i = 0} ^{k} C_{k} ^{i} S'_i(0) m^{k - i}\\ 两边对m从0->n求积分\\ S_k(n) - S_k(0) = \sum_{i = 0} ^{k} C_{k} ^{i} S'_i(0) \frac{n ^{k - i + 1}}{k - i + 1} - \sum_{i = 0} ^{k} C_{k} ^{i} S'_i(0) \frac{0 ^{k - i + 1}}{k - i + 1}\\ S_k(0) = 0, 对于最右侧i \neq k + 1,所以0 ^{k - i + 1} = 0 \\ S_k(n) = \sum_{i = 0} ^{k} C_{k} ^{i} S'_i(0) \frac{n ^{k - i + 1}}{k - i + 1} = \frac{1}{k + 1} \sum_{i = 0} ^{k} C_{k + 1} ^{i} S'_i(0) n ^{k + i - 1}\\
(0+m)k=i=0∑kCki0imk−i⋮(n−1+m)k=i=0∑kCki(n−1)imk−i左右两边同时相加Sk(n+m)−Sk(m)=i=0∑kCkiSi(n)mk−i左右两边对n求导Sk′(n+m)=i=0∑kCkiSi′(n)mk−in=0，代入Sk′(m)=i=0∑kCkiSi′(0)mk−i两边对m从0−>n求积分Sk(n)−Sk(0)=i=0∑kCkiSi′(0)k−i+1nk−i+1−i=0∑kCkiSi′(0)k−i+10k−i+1Sk(0)=0,对于最右侧i=k+1,所以0k−i+1=0Sk(n)=i=0∑kCkiSi′(0)k−i+1nk−i+1=k+11i=0∑kCk+1iSi′(0)nk+i−1
其实，这里的
S
i
(
0
)
S_i(0)
Si(0)就是伯努利数，为了方便书写，
B
i
=
S
i
′
(
0
)
B_i = S'_i(0)
Bi=Si′(0)。
我
们
先
将
n
=
1
代
入
上
式
S
k
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1
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=
1
k
+
1
∑
i
=
0
k
C
k
+
1
i
B
i
同
样
的
我
们
把
右
边
的
最
高
项
给
提
出
来
S
k
(
1
)
=
1
k
+
1
∑
i
=
0
k
−
1
C
k
+
1
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B
i
+
B
k
则
有
B
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−
S
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1
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=
−
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k
+
1
∑
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+
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i
B
i
特
殊
地
，
考
虑
k
=
0
,
有
S
0
(
1
)
=
1
,
所
以
B
0
=
1
k
≥
1
,
S
k
(
1
)
=
0
,
所
以
B
k
=
−
1
k
+
1
∑
i
=
0
k
−
1
C
k
+
1
i
B
i
我们先将n = 1代入上式\\ S_k(1) = \frac{1}{k + 1} \sum_{i = 0} ^{k} C_{k + 1} ^{i} B_i\\ 同样的我们把右边的最高项给提出来\\ S_k(1) = \frac{1}{k + 1} \sum_{i = 0} ^{k - 1} C_{k + 1} ^{i} B_i + B_k\\ 则有B_k - S_k(1) = - \frac{1}{k + 1} \sum_{i = 0} ^{k - 1} C_{k + 1} ^{i} B_i\\ 特殊地，考虑k = 0,有S_0(1) = 1, 所以B_0 = 1\\ k \geq 1, S_k(1) = 0, 所以B_k = - \frac{1}{k + 1} \sum_{i = 0} ^{k - 1} C_{k + 1} ^{i} B_i\\
我们先将n=1代入上式Sk(1)=k+11i=0∑kCk+1iBi同样的我们把右边的最高项给提出来Sk(1)=k+11i=0∑k−1Ck+1iBi+Bk则有Bk−Sk(1)=−k+11i=0∑k−1Ck+1iBi特殊地，考虑k=0,有S0(1)=1,所以B0=1k≥1,Sk(1)=0,所以Bk=−k+11i=0∑k−1Ck+1iBi
由此我们可以
O
(
k
2
)
O(k ^ 2)
O(k2)递推出伯努利数出来了。
F a u l h a b e r Faulhaber Faulhaber公式
S k ( x ) = 1 k + 1 ∑ i = 0 k C k + 1 i B i x k − i + 1 S_k(x) = \frac{1}{k + 1} \sum\limits_{i = 0} ^{k}C_{k + 1} ^{i}B_i x ^{k - i + 1} Sk(x)=k+11i=0∑kCk+1iBixk−i+1
一个自然数幂次求和的公式，其实就是我们推导过程中的一步。
#include <bits/stdc++.h>
using namespace std;
const int N = 3e3 +10, mod = 1e9 + 7;
int C[N][N], B[N], inv[N];
void init() {
for (int i = 0; i < N; i++) {
C[i][0] = C[i][i] = 1;
for (int j = 1; j < i; j++) {
C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
}
}
inv[1] = 1;
for (int i = 2; i < N; i++) {
inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
}
B[0] = 1;
for (int i = 1; i < N - 1; i++) {
int cur = 0;
for (int j = 0; j < i; j++) {
cur = (cur + 1ll * C[i + 1][j] * B[j] % mod) % mod;
}
cur = 1ll * cur * inv[i + 1] % mod;
B[i] = (mod - cur) % mod;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
init();
int T, k;
scanf("%d", &T);
while (T--) {
long long n, ans = 0, cur = 1;
scanf("%lld %d", &n, &k);
n++;
n %= mod;
for (int i = k; i >= 0; i--) {
cur = cur * n % mod;
ans = (ans + 1ll * C[k + 1][i] * B[i] % mod * cur % mod) % mod;
}
printf("%lld\n", ans * inv[k + 1] % mod);
}
return 0;
}
考虑生成函数
∑
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对
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加
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!
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=
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1
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+
B
n
n
!
设
指
数
型
生
成
函
数
B
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x
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=
∑
n
≥
0
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n
x
n
有
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从
而
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n
\sum_{i = 0} ^{n} C_{n + 1} ^{i} B_i = [n = 0]\\ 对两边加上B_{n + 1}\\ \sum_{i = 0} ^{n + 1} C_{n + 1} ^{i} B_i = [n = 0] + B_{n + 1}\\ \sum_{i = 0} ^{n} C_{n} ^{i} B_i = [n = 1] + B_{n}\\ \sum_{i = 0} ^{n} \frac{1}{i!(n - i)!} B_{i} = [n = 1] + \frac{B_{n}}{n!}\\ 设指数型生成函数B(x) = \sum_{n \geq 0} B_n x ^ n\\ 有B(x) e ^x = x + B(x)\\ 从而有B(x) = \frac{x}{e ^ x - 1}\\ B(x) = \frac{x}{\sum\limits_{n \geq 0} \frac{1}{n!} x ^ n - 1}\\ \frac{x}{\sum\limits_{n \geq 1} \frac{1}{n!} x ^ n}\\ \frac{1}{\sum\limits_{n \geq 0} \frac{1}{(n + 1)!}x ^ n}\\
i=0∑nCn+1iBi=[n=0]对两边加上Bn+1i=0∑n+1Cn+1iBi=[n=0]+Bn+1i=0∑nCniBi=[n=1]+Bni=0∑ni!(n−i)!1Bi=[n=1]+n!Bn设指数型生成函数B(x)=n≥0∑Bnxn有B(x)ex=x+B(x)从而有B(x)=ex−1xB(x)=n≥0∑n!1xn−1xn≥1∑n!1xnxn≥0∑(n+1)!1xn1
只需多项式求逆即可解决。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, mod = 998244353;
int r[N], b[N], t[N];
int quick_pow(int a, int n) {
int ans = 1;
while (n) {
if (n & 1) {
ans = 1ll * a * ans % mod;
}
a = 1ll * a * a % mod;
n >>= 1;
}
return ans;
}
void get_r(int lim) {
for (int i = 0; i < lim; i++) {
r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
}
}
void NTT(int *f, int lim, int rev) {
for (int i = 0; i < lim; i++) {
if (i < r[i]) {
swap(f[i], f[r[i]]);
}
}
for (int mid = 1; mid < lim; mid <<= 1) {
int wn = quick_pow(3, (mod - 1) / (mid << 1));
for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
int w = 1;
for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
}
}
}
if (rev == -1) {
int inv = quick_pow(lim, mod - 2);
reverse(f + 1, f + lim);
for (int i = 0; i < lim; i++) {
f[i] = 1ll * f[i] * inv % mod;
}
}
}
void polyinv(int *f, int *g, int n) {
if (n == 1) {
g[0] = quick_pow(f[0], mod - 2);
return ;
}
polyinv(f, g, n + 1 >> 1);
for (int i = 0; i < n; i++) {
t[i] = f[i];
}
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
get_r(lim);
NTT(t, lim, 1);
NTT(g, lim, 1);
for (int i = 0; i < lim; i++) {
int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;
g[i] = 1ll * g[i] * cur % mod;
t[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
}
int fac[N], ifac[N], B[N], n;
void init() {
fac[0] = 1;
for (int i = 1; i < N; i++) {
fac[i] = 1ll * fac[i - 1] * i % mod;
}
ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);
for (int i = N - 2; i >= 0; i--) {
ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
init();
n = 100000;
for (int i = 0; i <= n + 1; i++) {
B[i] = ifac[i + 1];
}
polyinv(B, b, n + 2);
for (int i = 0; i <= n + 1; i++) {
B[i] = 1ll * b[i] * fac[i] % mod;
}
for (int i = 0; i <= n + 1; i++) {
printf("%d%c", B[i], i == n + 1 ? '\n' : ' ');
}
return 0;
}
P3711 仓鼠的数学题
F
(
x
)
=
∑
k
=
0
n
S
k
(
x
)
a
k
原
本
定
义
的
S
k
(
x
)
=
∑
i
=
0
x
i
k
根
据
伯
努
利
数
的
定
义
S
k
′
(
x
)
=
∑
i
=
0
x
−
1
i
k
则
我
们
求
F
(
x
)
=
∑
k
=
0
n
S
k
′
(
x
)
a
k
,
答
案
即
为
F
(
x
+
1
)
考
虑
先
求
F
(
x
)
=
∑
k
=
0
n
a
k
1
k
+
1
∑
i
=
0
k
C
k
+
1
i
B
i
x
k
−
i
+
1
∑
k
=
0
n
a
k
k
+
1
∑
i
=
0
k
(
k
+
1
)
!
i
!
(
k
+
1
−
i
)
!
B
i
x
k
−
i
+
1
∑
k
=
0
n
a
k
k
!
∑
i
=
0
k
B
i
i
!
x
k
−
i
+
1
(
k
−
i
+
1
)
!
∑
i
=
1
n
+
1
x
i
i
!
∑
k
=
i
−
1
n
(
a
k
k
!
)
(
B
k
+
1
−
i
(
k
+
1
−
i
)
!
)
设
f
(
n
)
=
a
k
k
!
,
g
(
n
)
=
B
n
n
!
,
另
g
(
n
)
=
g
(
n
−
i
)
即
翻
转
一
下
∑
i
=
1
n
+
1
x
i
i
!
∑
k
=
i
−
1
n
f
(
k
)
g
(
n
+
i
−
k
)
容
易
发
现
后
面
是
一
个
卷
积
，
所
以
可
以
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F(x) = \sum_{k = 0} ^{n} S_k(x) a_k\\ 原本定义的S_k(x) = \sum_{i = 0} ^{x} i ^ k\\ 根据伯努利数的定义S'_k(x) = \sum_{i = 0} ^{x - 1} i ^ k\\ 则我们求F(x) = \sum_{k = 0} ^{n} S'_k(x) a_k, 答案即为F(x + 1)\\ 考虑先求F(x) = \sum_{k = 0} ^{n} a_k \frac{1}{k + 1} \sum\limits_{i = 0} ^{k}C_{k + 1} ^{i}B_i x ^{k - i + 1}\\ \sum_{k = 0} ^{n} \frac{a_k}{k + 1} \sum_{i = 0} ^{k} \frac{(k + 1)!}{i!(k + 1 - i)!} B_i x ^{k - i + 1}\\ \sum_{k = 0} ^{n} a_k k! \sum_{i = 0} ^{k} \frac{B_i}{i!}\frac{x ^{k - i + 1}}{(k - i + 1)!}\\ \sum_{i = 1} ^{n + 1} \frac{x ^{i}}{i!} \sum_{k = i - 1} ^{n} \left(a_k k!\right) \left( \frac{B_{k + 1 - i}}{(k + 1 - i)!} \right)\\ 设f(n) = a_k k!, g(n) = \frac{B_{n}}{n!}, 另g(n) = g(n - i)即翻转一下\\ \sum_{i = 1} ^{n + 1} \frac{x ^ i}{i!} \sum_{k = i - 1} ^{n} f(k) g(n + i - k)\\ 容易发现后面是一个卷积，所以可以优化\\
F(x)=k=0∑nSk(x)ak原本定义的Sk(x)=i=0∑xik根据伯努利数的定义Sk′(x)=i=0∑x−1ik则我们求F(x)=k=0∑nSk′(x)ak,答案即为F(x+1)考虑先求F(x)=k=0∑nakk+11i=0∑kCk+1iBixk−i+1k=0∑nk+1aki=0∑ki!(k+1−i)!(k+1)!Bixk−i+1k=0∑nakk!i=0∑ki!Bi(k−i+1)!xk−i+1i=1∑n+1i!xik=i−1∑n(akk!)((k+1−i)!Bk+1−i)设f(n)=akk!,g(n)=n!Bn,另g(n)=g(n−i)即翻转一下i=1∑n+1i!xik=i−1∑nf(k)g(n+i−k)容易发现后面是一个卷积，所以可以优化
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设F(x) = \sum_{i = 0} ^{n} a_i x ^ i\\ F(x + 1) = \sum_{i = 0} ^{n} a_i (x + 1) ^ i\\ \sum_{i = 0} ^{n} a_i \sum_{j = 0} ^{i} C_{j} ^{i} x ^ j\\ \sum_{i = 0} ^{n} a_i \sum_{j = 0} ^{i} \frac{i!}{j!(i - j)!} x ^ j\\ \sum_{i = 0} ^{n} \frac{x ^ i}{i!} \sum_{j = i} ^{n} a_j j! \frac{1}{(j - i)!}\\ 另f(n) = a_n n!, g(n) = \frac{1}{n!},再翻转g(n)\\ \sum_{i = 0} ^{n} \frac{x ^{i}}{i!} \sum_{j = i} ^{n} f(j) g(n + i - j)\\ 后面也同样是一个卷积，所以可以优化\\
设F(x)=i=0∑naixiF(x+1)=i=0∑nai(x+1)ii=0∑naij=0∑iCjixji=0∑naij=0∑ij!(i−j)!i!xji=0∑ni!xij=i∑najj!(j−i)!1另f(n)=ann!,g(n)=n!1,再翻转g(n)i=0∑ni!xij=i∑nf(j)g(n+i−j)后面也同样是一个卷积，所以可以优化
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, mod = 998244353;
int r[N], t[N];
int quick_pow(int a, int n) {
int ans = 1;
while (n) {
if (n & 1) {
ans = 1ll * a * ans % mod;
}
a = 1ll * a * a % mod;
n >>= 1;
}
return ans;
}
void get_r(int lim) {
for (int i = 0; i < lim; i++) {
r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
}
}
void NTT(int *f, int lim, int rev) {
for (int i = 0; i < lim; i++) {
if (i < r[i]) {
swap(f[i], f[r[i]]);
}
}
for (int mid = 1; mid < lim; mid <<= 1) {
int wn = quick_pow(3, (mod - 1) / (mid << 1));
for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
int w = 1;
for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
}
}
}
if (rev == -1) {
int inv = quick_pow(lim, mod - 2);
reverse(f + 1, f + lim);
for (int i = 0; i < lim; i++) {
f[i] = 1ll * f[i] * inv % mod;
}
}
}
void polyinv(int *f, int *g, int n) {
if (n == 1) {
g[0] = quick_pow(f[0], mod - 2);
return ;
}
polyinv(f, g, n + 1 >> 1);
for (int i = 0; i < n; i++) {
t[i] = f[i];
}
int lim = 1;
while (lim < 2 * n) {
lim <<= 1;
}
get_r(lim);
NTT(t, lim, 1);
NTT(g, lim, 1);
for (int i = 0; i < lim; i++) {
int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;
g[i] = 1ll * g[i] * cur % mod;
t[i] = 0;
}
NTT(g, lim, -1);
for (int i = n; i < lim; i++) {
g[i] = 0;
}
}
int fac[N], ifac[N], a[N], f[N], g[N], B[N], n;
void init() {
fac[0] = 1;
for (int i = 1; i < N; i++) {
fac[i] = 1ll * fac[i - 1] * i % mod;
}
ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);
for (int i = N - 2; i >= 0; i--) {
ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
init();
scanf("%d", &n);
for (int i = 0; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = 0; i <= n + 1; i++) {
B[i] = ifac[i + 1];
}
polyinv(B, g, n + 2);
for (int i = 0; i <= n; i++) {
f[i] = 1ll * a[i] * fac[i] % mod;
}
reverse(g, g + n + 2);
int lim = 1;
while (lim < 2 * n + 10) {
lim <<= 1;
}
get_r(lim);
NTT(f, lim, 1), NTT(g, lim, 1);
for (int i = 0; i < lim; i++) {
f[i] = 1ll * f[i] * g[i] % mod;
}
NTT(f, lim, -1);
for (int i = 1; i <= n + 1; i++) {
a[i] = 1ll * ifac[i] * f[n + i] % mod;
}
a[0] = 0;
n++;
for (int i = 0; i < lim; i++) {
f[i] = g[i] = 0;
}
for (int i = 0; i <= n; i++) {
f[i] = 1ll * a[i] * fac[i] % mod;
g[i] = ifac[n - i];
}
NTT(f, lim, 1), NTT(g, lim, 1);
for (int i = 0; i < lim; i++) {
f[i] = 1ll * f[i] * g[i] % mod;
}
NTT(f, lim, -1);
for (int i = 0; i <= n; i++) {
printf("%lld%c", 1ll * ifac[i] * f[n + i] % mod, i == n ? '\n' : ' ');
}
return 0;
}
标签：仓鼠,frac,int,lim,sum,1ll,P3711,例题,mod 来源： https://blog.csdn.net/weixin_45483201/article/details/114414524
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