标签：end 题解 return splay The div and pow log
The Happy Prince and Other Tales
给定长度为 \(n+1\) 的数列 \(\langle a_i\rangle_{i=0}^n\)，请对于 \(\forall m\in[0,n]\) 输出 \(f_m(n)\bmod 998244353\)，
\[f_m(n)=\sum_{i=0}^na_i\sum_{k=0}^ib^k\binom{m}{k}\binom{n-m}{i-k}
\]
\((n\le 10^6,a_i,b< 998244353)\)
我们令
\[a_i=\sum_{j_1=0}^i\sum_{j_2=0}^{j_1}\cdots\sum_{j_p=0}^{j_{p-1}}\binom{i}{j_1}\binom{j_1}{j_2}\cdots\binom{j_{p-1}}{j_p}c_{j_p}
\]
则
\[\begin{aligned}
f_m(n)=&\sum_{i=0}^n\sum_{j_1=0}^i\sum_{j_2=0}^{j_1}\cdots\sum_{j_p=0}^{j_{p-1}}\binom{i}{j_1}\binom{j_1}{j_2}\cdots\binom{j_{p-1}}{j_p}c_{j_p}\sum_{k=0}^ib^k\binom{m}{k}\binom{n-m}{i-k} \=&\sum_{j_p=0}^nc_{j_p}\sum_{i=0}^n\sum_{j_1=0}^i\sum_{j_2=0}^{j_1}\cdots\sum_{j_{p-1}=j_p}^{j_{p-2}}\sum_{k=0}^ib^k\binom{m}{k}\binom{n-m}{i-k}\binom{i}{j_1}\binom{j_1}{j_2}\cdots\binom{j_{p-1}}{j_p} \\end{aligned}
\]
存在
\[\begin{aligned}
&\sum_{i_1=0}^n\sum_{i_2=0}^{i_1}\cdots\sum_{i_p=0}^{i_{p-1}}\binom{n}{i_1}\binom{i_1}{i_2}\cdots\binom{i_{p-1}}{i_p}\binom{i_p}{t}k^{i_1} \=&\sum_{i_1=0}^n\sum_{i_2=0}^{i_1}\cdots\sum_{i_p=0}^{i_{p-1}}\frac{n!}{(n-i_1)!(i_1-i_2)!\cdots(i_{p-1}-i_p)!(i_p-t)!t!}k^{i_1} \=&\sum_{\sum_{j=1}^pi_j=n-t}\frac{n!}{t!\prod_{j=1}^pi_j!}k^{n-i_1} \\end{aligned}
\]
发现这是一 个多项式定理的形式，考虑生成函数，上式等于
\[[x^t](kp+kx+1)^n
\]
重新观察 \(f_m(n)\)，可以进行如下化简
\[\begin{aligned}
f_m(n)=&\sum_{j_{p}=0}^nc_{j_p}\sum_{k=0}[x^k](1+bp+bx)^m[x^{j_p-k}](1+p+x)^{n-m} \=&\sum_{j_{p}=0}^nc_{j_p}[x^{j_p}](1+bp+bx)^m(1+p+x)^{n-m}
\end{aligned}
\]
我们令 \(p=-1\)，带入得
\[f_m(n)=\sum_{i=n-m}^nc_i[x^{i-(n-m)}](1-b+bx)^m=\sum_{i=n-m}^nc_i\binom{m}{i-(n-m)}b^{i-(n-m)}(1-b)^{n-i}
\]
这是多项式卷积的形式，即
\[\frac{f_m(n)}{m!}=\sum_{i}\frac{b^{m-(n-i)}}{(m-(n-i))!}\cdot\frac{(1-b)^{n-i}c_i}{(n-i)!}
\]
我们现在来考虑如何求 \(c\)，根据反演，有
\[c_i=\sum_{j_1=0}^i\sum_{j_2=0}^{j_1}\cdots\sum_{j_p=0}^{j_{p-1}}\binom{i}{j_1}\binom{j_1}{j_2}\cdots\binom{j_{p-1}}{j_p}(-1)^{i-j_1}(-1)^{j_1-j_2}\cdots(-1)^{j_{p-1}-j_p}a_{j_p}
\]
发现这个仍是多项式定理的形式，即
\[c_i=\sum_{j=0}^i[x^j](-p+\sqrt[j]{a_j})^i=\sum_{j=0}^i\binom{i}{j}(-p)^{i-j}a_j=\sum_{j=0}^i\binom{i}{j}a_j
\]
至此，共两次多项式卷积即可求出答案，时间复杂度 \(O(n\log n)\)。
#include <iostream>
using i64 = long long;
namespace fio;
namespace ply;
using ply::MOD;
using ply::fac;
using ply::ifc;
using ply::pow;
const int N = 1000000 + 7;
int n;
i64 a[N], b, c[N];
int main() {
fio::in(n), fio::in(b), ++n;
for (int i = 0; i < n; ++i) fio::in(a[i]);
static ply::polyT f, g, h;
for (int i = 0; i < n; ++i)
f[i] = ifc[i], g[i] = a[i] * ifc[i] % MOD;
ply::mul(f, n, g, n, h);
for (int i = 0; i < n; ++i)
c[i] = h[i] * fac[i] % MOD;
for (int i = 0; i < n; ++i)
f[i] = pow(b, i) * ifc[i] % MOD,
g[i] = pow(1 - b, i) * c[n - i - 1] % MOD * ifc[i] % MOD;
ply::mul(f, n, g, n, h);
for (int i = 0; i < n; ++i)
fio::out((h[i] * fac[i] % MOD + MOD) % MOD), fio::put(‘\n‘);
return 0;
}
FZOJ4167 The Happy Prince and Other Tales 题解
标签：end 题解 return splay The div and pow log
原文地址：https://www.cnblogs.com/Ryedii-blog/p/13024104.html
