LibreOJ 3315.「ZJOI2020」抽卡

继续数数。

考虑计算 \(i\) 步以后还没有连续 \(k\) 张卡的概率。
容易发现可以转化为对于所有没有连续 \(k\) 张卡的状态,计算 \(i\) 步后到达该状态的概率并求和。

\(f_i\) 表示从 \(m\) 张卡里选择 \(i\) 张卡使得没有连续 \(k\) 张卡的方案数。
则答案即 \[ \begin{aligned} & \sum\limits_{i=0}^{\infty} \sum\limits_{j=0}^{m-1} f_j \sum\limits_{u=0}^j \binom ju (-1)^{j-u} \left(\frac um\right)^i \\ = & \sum\limits_{j=0}^{m-1} f_j \sum\limits_{u=0}^j \binom ju (-1)^{j-u} \sum\limits_{i=0}^{\infty} \left(\frac um\right)^i \\ = & \sum\limits_{j=0}^{m-1} f_j \sum\limits_{u=0}^j \binom ju (-1)^{j-u} \frac m{m-u} \end{aligned} \]

后面的和式可以随便卷积一下计算,那么问题转化为计算 \(f_i\)
容易发现实际上可以将序列 \(a\) 排序后划分为若干极大连续段,对于每个极大连续段分别计算答案再分治 NTT 卷起来。

那么对于一个长度为 \(n\) 的连续段,考虑计算其中的 \(f_i\)
略加思考后不难发现这个东西就是 \[ [x^{n+1}] \left(\sum\limits_{j=1}^k x^j\right)^{n-i+1} \]

即,将 \(n\) 张卡的状态视作 \(0/1\);对于每个 \(0\),将其与之前最长的连续 \(1\)(可以为空)合并在一起考虑;那么每一个这样的块的长度便是 \(1\dots k\)
此外,需要在最后强制加入一个 \(0\)

那么设 \(G(x) = \frac{x-x^{k+1}}{1-x}\),不难发现,若写出 \(f\) 的生成函数,则有 \[ \sum\limits_{i=0}^n f_i z^i = [x^{n+1}] \frac1{1-G(x)z} \]

考虑对这样一个二元生成函数应用扩展拉格朗日反演,有 \[ [x^{n+1}] \frac1{1-G(x)z} = \frac 1{n+1}[x^n] \frac z{(1-xz)^2} \left(\frac x{G^{-1}(x)}\right)^{n+1} \]

其中 \(G^{-1}\)\(G\) 的复合逆。
若求出 \(G^{-1}\),结合多项式求逆和多项式快速幂并展开 \(\frac z{(1-xz)^2}\) 便可以求得 \(f_i\)
于是问题变为求解 \(G^{-1}(x)\)

\(F = G^{-1}\),考虑牛顿迭代之。
设已得 \[ G(F_0(x)) \equiv x \pmod{x^n} \]

\(F(x)=F_0(x)\) 处泰勒展开 \[ \begin{aligned} G(F(x)) &\equiv G(F_0(x)) + G'(F_0(x))(F(x) - F_0(x)) \pmod{x^{2n}} \\ F(x) &\equiv F_0(x) + \frac{x-G(F_0(x))}{G'(F_0(x))} \pmod{x^{2n}} \end{aligned} \]

而易得 \[ G'(x) = \frac{kx^{k+1}-(k+1)x^k+1}{(1-x)^2} \]

故倍增 + 多项式全家桶即可(雾
此部分仅有一个 log,算上分治 NTT 总复杂度 \(O(m \log^2 m)\)
(当然这一个 log 顶人家不知道多少次分治 NTT)

貌似有更优美做法,懒得学了(
生成函数暴力美学(

代码:

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#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#define add(a,b) (a + b >= mod ? a + b - mod : a + b)
#define dec(a,b) (a < b ? a - b + mod : a - b)
using namespace std;
const int M = 2e5;
const int mod = 998244353;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
int n,m,K;
int a[M + 5];
int len[M + 5],tot;
namespace Poly
{
const int N = 1 << 19;
const int G = 3;
int lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init()
{
for(register int i = 2;i <= N;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / N);
rt[N >> 1] = 1;
for(register int i = (N >> 1) + 1;i <= N;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (N >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= N;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[N] = fpow(fac[N],mod - 2);
for(register int i = N;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= N;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
struct poly
{
vector<int> a;
inline poly(int x = 0)
{
x && (a.push_back(x),1);
}
inline poly(const vector<int> &o)
{
a = o,shrink();
}
inline poly(const poly &o)
{
a = o.a,shrink();
}
inline void shrink()
{
for(;!a.empty() && !a.back();a.pop_back());
}
inline int size() const
{
return a.size();
}
inline void resize(int x)
{
a.resize(x);
}
inline int operator[](int x) const
{
if(x < 0 || x >= size())
return 0;
return a[x];
}
inline void clear()
{
vector<int>().swap(a);
}
inline void ntt(int type = 1)
{
int n = size();
type == -1 && (reverse(a.begin() + 1,a.end()),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
friend inline poly operator+(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = add(a[i],b[i]);
return poly(ret);
}
friend inline poly operator-(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = dec(a[i],b[i]);
return poly(ret);
}
friend inline poly operator*(poly a,poly b)
{
if(a.a.empty() || b.a.empty())
return poly();
int lim = 1,tot = a.size() + b.size() - 1;
for(;lim < tot;lim <<= 1);
a.resize(lim),b.resize(lim);
a.ntt(),b.ntt();
for(register int i = 0;i < lim;++i)
a.a[i] = (long long)a[i] * b[i] % mod;
a.ntt(-1),a.shrink();
return a;
}
poly &operator+=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = add(a[i],o[i]);
return *this;
}
poly &operator-=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = dec(a[i],o[i]);
return *this;
}
poly &operator*=(poly o)
{
return (*this) = (*this) * o;
}
poly deriv() const
{
if(a.empty())
return poly();
vector<int> ret(size() - 1);
for(register int i = 0;i < size() - 1;++i)
ret[i] = (long long)(i + 1) * a[i + 1] % mod;
return poly(ret);
}
poly integ() const
{
if(a.empty())
return poly();
vector<int> ret(size() + 1);
for(register int i = 0;i < size();++i)
ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
return poly(ret);
}
inline poly modxn(int n) const
{
if(a.empty())
return poly();
n = min(n,size());
return poly(vector<int>(a.begin(),a.begin() + n));
}
inline poly inver(int m) const
{
poly ret(fpow(a[0],mod - 2));
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
return ret.modxn(m);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ().modxn(m);
}
inline poly exp(int m) const
{
poly ret(1);
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
return ret.modxn(m);
}
inline poly pow(int m,int k1,int k2 = -1) const
{
if(a.empty())
return poly();
if(k2 == -1)
k2 = k1;
int t = 0;
for(;t < size() && !a[t];++t);
if((long long)t * k1 >= m)
return poly();
poly ret;
ret.resize(m);
int u = fpow(a[t],mod - 2),v = fpow(a[t],k2);
for(register int i = 0;i < m - t * k1;++i)
ret.a[i] = (long long)operator[](i + t) * u % mod;
ret = ret.log(m - t * k1);
for(register int i = 0;i < ret.size();++i)
ret.a[i] = (long long)ret[i] * k1 % mod;
ret = ret.exp(m - t * k1),t *= k1,ret.resize(m);
for(register int i = m - 1;i >= t;--i)
ret.a[i] = (long long)ret[i - t] * v % mod;
for(register int i = 0;i < t;++i)
ret.a[i] = 0;
return ret;
}
};
}
using Poly::init;
using Poly::poly;
poly f,g[M + 5],h1,h2;
int ans;
inline poly calc(int m)
{
poly ret,powk,powk1,g,d;
for(register int k = 1;k < m;)
k <<= 1,
powk1 = ((powk = ret.pow(k,K)) * ret).modxn(k),
g.resize(2),g.a[0] = 0,g.a[1] = 1,
g -= ((1 - ret).inver(k) * (ret - powk1)).modxn(k),
d = (((1 - ret) * (1 - ret)).inver(k) * (powk1 * K - powk * (K + 1) + 1)).modxn(k),
ret = (ret + g * d.inver(k)).modxn(k);
return ret.modxn(m);
}
poly solve(int l,int r)
{
if(l == r)
return g[l];
int mid = l + r >> 1;
return solve(l,mid) * solve(mid + 1,r);
}
int main()
{
init();
scanf("%d%d",&m,&K),f = calc(m + 2);
for(register int i = 0;i < f.size() - 1;++i)
f.a[i] = f[i + 1];
f.resize(f.size() - 1),f = f.inver(m + 1);
for(register int i = 1;i <= m;++i)
scanf("%d",a + i);
sort(a + 1,a + m + 1),a[0] = -1;
for(register int i = 1;i <= m;++i)
a[i] == a[i - 1] + 1 ? (++len[tot]) : (len[++tot] = 1);
for(register int i = 1;i <= tot;++i)
{
poly t = f.pow(len[i] + 1,len[i] + 1);
g[i].resize(len[i] + 1);
for(register int j = 1;j <= len[i];++j)
g[i].a[len[i] - j + 1] = (long long)Poly::inv[len[i] + 1] * j % mod * t[len[i] - j + 1] % mod;
g[i].a[0] = 1;
}
f = solve(1,tot).modxn(m),h1.resize(m),h2.resize(m);
for(register int i = 0;i < m;++i)
h1.a[i] = (long long)(i & 1 ? mod - 1 : 1) * Poly::ifac[i] % mod,
h2.a[i] = (long long)m * Poly::inv[m - i] % mod * Poly::ifac[i] % mod;
h1 = (h1 * h2).modxn(m);
for(register int i = 0;i < m;++i)
ans = (ans + (long long)f[i] * h1[i] % mod * Poly::fac[i] % mod) % mod;
printf("%d\n",ans);
}