UOJ 593.新年的军队

新年的军队题解人话重置版!

首先作一步相当重要的转化:
对所有满足恰有 \(m\) 个位置 \(i\) 满足 \(\alpha_i > \alpha_{i+1}\) 的实数序列 \(\alpha_1,\alpha_2,\dots,\alpha_n\),计算 \(\alpha_k\) 的排名的分布。其中 \(\alpha_i\)\([0,1]\) 上均匀分布。

我们知道若干随机实数相等的概率是 \(0\),所以这里其实区间的开闭也不影响结果。那么若干个实数就可以排出排名,并且显然每种排列出现的概率是相等的。
所以这个问题的结果和原问题是等价的。

然后稍微插入一下对概率密度函数的普及。
(我为啥要开个引用块?为了让会的神仙方便区分我的瞎扯和正经内容)
假装我们有一个在 \([0,4]\) 上均匀分布的实数变量,考虑一个非常憨批的问题:其落在区间 \([1,2]\) 内的概率。
我们发现这就是长度之比的事。
不过这个时候我们假装有一个在函数图像上的长方形,横坐标的范围是 \([0,4]\),纵坐标都是 \(\frac14\),然后我们在 \([1,2]\) 上作定积分就得到了它的概率。
这个 \(\frac14\) 其实就是个类似概率密度函数的东西。它用来描述连续型随机变量的输出值。我们想要知道它落在某个区间上的概率,只需要在这个区间上对概率密度函数作积分即可。

我们考虑用概率密度来考察特定 \(\alpha_k\) 的分布。若 \(\alpha_k\) 的排名为 \(j\),则其贡献的概率密度即为 \[ \frac{ x^{j-1}(1-x)^{n-j} }{ (j-1)!(n-j)! } \]

这是因为当 \(\alpha_k = x\) 时,有 \(j-1\) 个数小于其的概率各是 \(x\),然后它们有 \((j-1)!\) 种排列方法,大于其的情况类似。
这个地方我们需要考虑除以 \((j-1)!\) 的原因在于这些情况是本质相同的。

这样的话,刻画两个数之间的大小关系就变得容易了。因为我们可以直接在 \([0,x]\)\([x,1]\) 上积分。

接下来我们建立所求的东西和概率密度函数的关系。
考虑令 \(a_i\)\(p_i=i+1\) 的方案数,设 \(f(x)\)\(\alpha_k\) 最终的概率密度函数,那么我们有 \[ f(x) = \sum\limits_{i=0}^{n-1} a_i \frac{ x^i (1-x)^{n-1-i} }{i!(n-1-i)!} \]

容易反演 \[ \sum\limits_{i=0}^{n-1} a_i \frac{x^i}{i!(n-1-i)!} = \sum\limits_{i=0}^{n-1} f_i x^i (1+x)^{n-1-i} \]

具体地 \[ \begin{aligned} \sum\limits_{i=0}^{n-1} f_i x^i (1+x)^{n-1-i} &= \sum\limits_{i=0}^{n-1} (1+x)^i f_{n-1-i} x^{n-1-i} \\ &= \sum\limits_{i=0}^{n-1} f_{n-1-i} \sum\limits_{j=0}^i \binom ij x^{n-1-(i-j)} \\ &= \sum\limits_{i=0}^{n-1} f_{n-1-i} \sum\limits_{j=0}^i \binom ij x^{n-1-j} \\ &= \sum\limits_{j=0}^{n-1} x^{n-1-j} \sum\limits_{i=j}^{n-1} \binom ij f_{n-1-i} \end{aligned} \]

接下来正式开始处理原问题。
考虑在概率密度函数中加入两个元区分大于号和小于号:设 \(F(u,v,t)\) 表示排列末尾的分布,其中 \(u\) 记录小于号,\(v\) 记录大于号。 \[ \def\d{ {\rm d} } \def\e{ {\rm e} } \def\D{ {\rm D} } F(t) = 1 + u \int_0^t F(\tau) \d \tau + v \int_t^1 F(\tau) \d \tau \]

因此 \[ F' = (u - v)F \]

显然可以待定系数得到其解 \[ F = C\e^{(u-v)t} \]

代入 \(t=0,1\) 可以确定 \[ F = \frac{ (u - v)\e^{(u - v)t} }{ u - v\e^{u-v} } \]

接下来用 \(x\) 元计量边数,用 \(y\) 元计量大于号的数量。
对于 \(p_k\) 的左侧,我们令 \(u=x,v=xy\);对于右侧我们令 \(u=xy,v=x\)。那么 \[ \begin{aligned} L &= \frac{ (1-y)\e^{x(1-y)t} }{ 1-y\e^{x(1-y)} } \\ R &= \frac{ (1-y)\e^{x(1-y)(1-t)} }{ 1-y\e^{x(1-y)} } \end{aligned} \]

接下来有两条路。

离散微积分

我们直接考虑提取其系数。
我们知道 \(([x^{n_1}] L) ([x^{n_2} R])\) 就等于 \[ \frac1{n_1!n_2!} (1-y)^{n_1+n_2+2}\left(\sum\limits_{j\ge 0} y^j (t+j)^{n_1}\right) \left(\sum\limits_{j\ge 0} y^j ((j+1)-t)^{n_2}\right) \]

事实上在这个地方我们容易求出结果(关于 \(t\) 的多项式)在 \(0,1,\dots,n\) 处的点值,能否通过取决于插值的常数。
不过这个做法不太有趣,此处略过。

去掉常数 \(\frac{(-1)^{n_2}}{n_1!n_2!}\),我们把它变成 \[ \begin{aligned} & \quad\; [y^m] (1-y)^{n+1} \sum\limits_{i\ge 0} \sum\limits_{j\ge 0} y^{i+j} (t+i)^{n_1} (t-(j+1))^{n_2} \\ &= \sum\limits_{s=0}^m [y^m](1-y)^{n+1} y^s \sum\limits_{i=0}^s (t+i)^{n_1} (t+i-(s+1))^{n_2} \end{aligned} \]

我们将其拆成正方向的无穷求和加上一个差分 \(\Delta_{s+1} f(t) = f(t) - f(t+s+1)\)
\(f_s = [y^m](1-y)^{n+1} y^s\)
\[ \sum\limits_{s=0}^m f_s \Delta_{s+1} \left(\sum\limits_{i\ge 0} (t+i)^{n_1} (t+i-(s+1))^{n_2}\right) \]

这个地方我们考虑把差分换到无穷求和里面,看起来我们求和的东西是不变的,但是实际上无穷求和各项的关系并不这么单纯,正像你作不定积分后会多出一个常数。
所以实际上变成了 \[ \sum\limits_{s=0}^m f_s \left(\sum\limits_{i\ge 0} \Delta_{s+1} (t+i)^{n_1} (t+i-(s+1))^{n_2}\right) + C \]

不管常数项,即 \[ \sum\limits_{i\ge 0} \left( \sum\limits_{s=0}^m f_s [(t+i)^{n_1} (t+i-(s+1))^{n_2}-(t+i+(s+1))^{n_1} (t+i)^{n_2}]\right) \]

但是这个 \(C\) 一看就不像是人工分析可以搞出来的东西……
不过回到原先的概率密度函数,我们会发现令 \(f_0\) 加上 \(C\) 会导致所有 \(a_i\) 之和加上 \(C\) 倍的某个常数(而这个常数是容易算出的),而我们知道 \(a_i\) 之和应当等于 \(\genfrac{\langle}{\rangle}{0pt}{}nm\)

\(\D\) 是求导算子,我们知道 \(f(t+c) = \e^{c \D} f(t)\),因而不妨设 \(F(x) = \sum\limits_{s=0}^m f_s x^{s+1}\),就有 \[ \sum\limits_{i\ge 0} \left((t+i)^{n_1} F(\e^{-\D}) (t+i)^{n_2}-(t+i)^{n_2} F(\e^{\D}) (t+i)^{n_1}\right) \]

我们知道我们作一次位移是 \(\e^{\D}\),正向差分是 \(1-\e^{\D}\),它的逆运算就是 \(\frac1{ 1-\e^{\D} }\),然后为了能够求逆我们恰好就凑出来了一个伯努利数的形式 \[ \int B(\D) \left((t+i)^{n_1} F(\e^{-\D}) (t+i)^{n_2}-(t+i)^{n_2} F(\e^{\D}) (t+i)^{n_1}\right), \quad B(t) = \frac{t}{1-\e^t} \]

接下来问题落在计算 \(G(x)\)。我们知道 \[ F(x) = \sum\limits_{i=0}^m \binom{n+1}{m-i} (-1)^{m-i} x^{i+1} \]

不妨设 \(F = xf\),令 \(C = (n-m+1)f_0 = (n-m+1) \binom{n+1}m (-1)^m\),那么我们知道 \[ \begin{aligned} (n-m+1+i)f_i &= -(m-i+1)f_{i-1} + [i=0]C \\ (n-m+1)f + xf' &= -mxf + x^2f' + C \end{aligned} \]

\(G(x) = F(\e^x) = \e^x f(\e^x) = \e^x g(x)\),那么 \[ \begin{aligned} (n-m+1)f(\e^x) + \e^x f'(\e^x) &= -m\e^xf(\e^x) + \e^{2x}f'(\e^x) + C \\ (n-m+1)g + g' &= -m \e^x g + \e^x g' + C \\ (n-m+1+m\e^x)g + (1-\e^x)g' &= C \\ ((n-m)\e^{-x}+m+1)G + (\e^{-x}-1)G' &= C \end{aligned} \]

提取 \([x^i]\)\[ \begin{aligned} (n-m)\sum\limits_{j=1}^i \frac{(-1)^j}{j!} G_{i-j} &+ (n+1) G_i \\ +\sum\limits_{j=1}^i \frac{(-1)^j}{j!} (i-j+1) G_{i-j+1} &= [i=0]C \end{aligned} \]

整理得 \[ \begin{aligned} (n-i+1)G_i &= [i=0]C \\ -(n-m)\sum\limits_{j=0}^{i-1} G_j \frac{ (-1)^{i-j} }{(i-j)!} &-\sum\limits_{j=1}^{i-1} j G_j \frac{ (-1)^{i-j+1} }{(i-j+1)!} \end{aligned} \]

四元生成函数

我们先不提取系数,而是先综合考虑整体的生成函数。

左侧和右侧的 \(x\) 元分别用 \(p,q\) 代替,令 \(n_1 = k-1,n_2 = n-k\),那么问题变为提取 \[ [y^m p^{n_1} q^{n_2}] (1-y)^2 \frac{ \e^{p(1-y)t + q(1-y)(1-t)} }{(1-y\e^{p(1-y)})(1-y\e^{q(1-y)})} \]

\(p,q\) 换元,提出 \((1-y)\),则 \[ (1-y)^{n+1} \frac{ \e^{pt + q(1-t)} }{(1-y\e^p)(1-y\e^q)} \]

作部分分式,则 \[ (1-y)^{n+1} \e^{pt+q(1-t)} \frac 1{\e^p - \e^q}\left(\frac{\e^p}{1-y\e^p} - \frac{\e^q}{1-y\e^q}\right) \]

这看起来不太能计算,但是某些奇思妙想让我们试图对 \(t\) 求导, \[ \begin{aligned} & \quad\; (1-y)^{n+1} \e^{pt+q(1-t)} \frac{p-q}{\e^p-\e^q}\left(\frac{\e^p}{1-y\e^p} - \frac{\e^q}{1-y\e^q}\right) \\ & = (1-y)^{n+1} \e^{pt+q(1-t)} \frac{ (p-q)\e^{-q} }{\e^{p-q}-1}\left(\frac{\e^p}{1-y\e^p} - \frac{\e^q}{1-y\e^q}\right) \\ & = (1-y)^{n+1} \e^{(p-q)t} B(p-q) \left(\frac{\e^p}{1-y\e^p} - \frac{\e^q}{1-y\e^q}\right) \end{aligned} \]

其中 \(B(x) = \frac x{\e^x-1}\)

损失的常数可以使用上面提到的方法类似处理。

\(F(x) = \sum\limits_{i\ge 0} [y^{m-i}] (1-y)^{n+1} x^{i+1},G(x) = F(\e^x)\),提取 \(y^m\) 的系数 \[ \e^{(p-q)t} B(p-q) (G(p)-G(q)) \]

分别考虑 \(\e^{(p-q)t} B(p-q) G(p),\e^{(p-q)t} B(p-q) G(q)\)
以前者为例,我们考虑换元:令 \(pr = q\),则变为 \[ \e^{p(1-r)t} B(p(1-r)) G(p) \]

对其提取 \(p^{n-1} r^{n_2}\) 的系数,有 \[ \begin{aligned} & \quad\; [p^{n-1} r^{n_2}] \e^{p(1-r)t} B(p(1-r)) G(p) \\ & = \sum\limits_{j \ge 0} ([p^j]B(p) \e^{pt})([r^{n_2}](1-r)^j)([p^{n-1-j}]G(p)) \end{aligned} \]

\(P(p) = \sum\limits_{j\ge 0} p^{n-1-j} ([r^{n_2}](1-r)^j)([p^{n-1-j}]G(p))\),那么 \[ [p^{n-1}] P(p) B(p) \e^{pt} = \sum\limits_{j \ge 0} \frac{t^j}{j!} [p^{n-1-j}] P(p) B(p) \]

类似地,设 \(Q(q) = \sum\limits_{j\ge 0} q^{n-1-j} ([r^{n_1}](r-1)^j)([q^{n-1-j}] G(q))\),就有 \[ \sum\limits_{j\ge 0} \frac{t^j}{j!} [q^{n-1-j}] Q(q) B(q) \]

\(G\) 的求算已经在上文提到过。

时间复杂度 \(O(n \log^2 n)\)\(O\left(\frac{n \log^2 n}{\log \log n}\right)\)

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 N = 5e5;
const int mod = 998244353;
int n,m,k;
int n1,n2;
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;
}
namespace Poly
{
const int LG = 19;
const int N = 1 << LG + 1;
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];
inline void init()
{
for(register int i = 2;i <= N;++i)
lg2[i] = lg2[i >> 1] + 1;
rt[0] = 1,rt[1 << LG] = fpow(G,(mod - 1) >> LG + 2);
for(register int i = LG;i;--i)
rt[1 << i - 1] = (long long)rt[1 << i] * rt[1 << i] % mod;
for(register int i = 1;i < N;++i)
rt[i] = (long long)rt[i & i - 1] * rt[i & -i] % mod;
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 poly rever() const
{
return poly(vector<int>(a.rbegin(),a.rend()));
}
inline void dif()
{
int n = size();
for(register int i = 0,len = n >> 1;len;++i,len >>= 1)
for(register int j = 0,*w = rt;j < n;j += len << 1,++w)
for(register int k = j,R;k < j + len;++k)
R = (long long)*w * a[k + len] % mod,
a[k + len] = dec(a[k],R),
a[k] = add(a[k],R);
}
inline void dit()
{
int n = size();
for(register int i = 0,len = 1;len < n;++i,len <<= 1)
for(register int j = 0,*w = rt;j < n;j += len << 1,++w)
for(register int k = j,R;k < j + len;++k)
R = add(a[k],a[k + len]),
a[k + len] = (long long)(a[k] - a[k + len] + mod) * *w % mod,
a[k] = R;
reverse(a.begin() + 1,a.end());
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
inline void ntt(int type = 1)
{
type == 1 ? dif() : dit();
}
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();
if(a.size() < 40 || b.size() < 40)
{
if(a.size() > b.size())
swap(a,b);
poly ret;
ret.resize(a.size() + b.size() - 1);
for(register int i = 0;i < ret.size();++i)
for(register int j = 0;j <= i && j < a.size();++j)
ret.a[i] = (ret[i] + (long long)a[j] * b[i - j]) % mod;
ret.shrink();
return ret;
}
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)),f,g;
for(register int k = 1;k < m;)
{
k <<= 1,f.resize(k),g.resize(k);
for(register int i = 0;i < k;++i)
f.a[i] = (*this)[i],g.a[i] = ret[i];
f.ntt(),g.ntt();
for(register int i = 0;i < k;++i)
f.a[i] = (long long)f[i] * g[i] % mod;
f.ntt(-1);
for(register int i = 0;i < (k >> 1);++i)
f.a[i] = 0;
f.ntt();
for(register int i = 0;i < k;++i)
f.a[i] = (long long)f[i] * g[i] % mod;
f.ntt(-1);
ret.resize(k);
for(register int i = (k >> 1);i < k;++i)
ret.a[i] = dec(0,f[i]);
}
return ret.modxn(m);
}
inline pair<poly,poly> div(poly o) const
{
if(size() < o.size())
return make_pair(poly(),*this);
poly f,g;
f = (rever().modxn(size() - o.size() + 1) * o.rever().inver(size() - o.size() + 1)).modxn(size() - o.size() + 1).rever();
g = (modxn(o.size() - 1) - o.modxn(o.size() - 1) * f.modxn(o.size() - 1)).modxn(o.size() - 1);
return make_pair(f,g);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ().modxn(m);
}
inline poly exp(int m) const
{
poly ret(1),iv,it,d = deriv(),itd,itd0,t1;
if(m < 70)
{
ret.resize(m);
for(register int i = 1;i < m;++i)
{
for(register int j = 1;j <= i;++j)
ret.a[i] = (ret[i] + (long long)j * operator[](j) % mod * ret[i - j]) % mod;
ret.a[i] = (long long)ret[i] * inv[i] % mod;
}
return ret;
}
for(register int k = 1;k < m;)
{
k <<= 1;
it.resize(k >> 1);
for(register int i = 0;i < (k >> 1);++i)
it.a[i] = ret[i];
itd = it.deriv(),itd.resize(k >> 1);
iv = ret.inver(k >> 1),iv.resize(k >> 1);
it.ntt(),itd.ntt(),iv.ntt();
for(register int i = 0;i < (k >> 1);++i)
it.a[i] = (long long)it[i] * iv[i] % mod,
itd.a[i] = (long long)itd[i] * iv[i] % mod;
it.ntt(-1),itd.ntt(-1),it.a[0] = dec(it[0],1);
for(register int i = 0;i < k - 1;++i)
itd.a[i % (k >> 1)] = dec(itd[i % (k >> 1)],d[i]);
itd0.resize((k >> 1) - 1);
for(register int i = 0;i < (k >> 1) - 1;++i)
itd0.a[i] = d[i];
itd0 = (itd0 * it).modxn((k >> 1) - 1);
t1.resize(k - 1);
for(register int i = (k >> 1) - 1;i < k - 1;++i)
t1.a[i] = itd[(i + (k >> 1)) % (k >> 1)];
for(register int i = k >> 1;i < k - 1;++i)
t1.a[i] = dec(t1[i],itd0[i - (k >> 1)]);
t1 = t1.integ();
for(register int i = 0;i < (k >> 1);++i)
t1.a[i] = t1[i + (k >> 1)];
for(register int i = (k >> 1);i < k;++i)
t1.a[i] = 0;
t1.resize(k >> 1),t1 = (t1 * ret).modxn(k >> 1),t1.resize(k);
for(register int i = (k >> 1);i < k;++i)
t1.a[i] = t1[i - (k >> 1)];
for(register int i = 0;i < (k >> 1);++i)
t1.a[i] = 0;
ret -= t1;
}
return ret.modxn(m);
}
inline poly sqrt(int m) const
{
poly ret(1),f,g;
for(register int k = 1;k < m;)
{
k <<= 1;
f = ret,f.resize(k >> 1);
f.ntt();
for(register int i = 0;i < (k >> 1);++i)
f.a[i] = (long long)f[i] * f[i] % mod;
f.ntt(-1);
for(register int i = 0;i < k;++i)
f.a[i % (k >> 1)] = dec(f[i % (k >> 1)],(*this)[i]);
g = (2 * ret).inver(k >> 1),f = (f * g).modxn(k >> 1),f.resize(k);
for(register int i = (k >> 1);i < k;++i)
f.a[i] = f[i - (k >> 1)];
for(register int i = 0;i < (k >> 1);++i)
f.a[i] = 0;
ret -= f;
}
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;
inline int C(int n,int m)
{
return n < m ? 0 : (long long)Poly::fac[n] * Poly::ifac[m] % mod * Poly::ifac[n - m] % mod;
}
poly f,b,g,p,q;
int rest,c;
void solve(int l,int r)
{
if(l == r)
{
g.a[l] = (long long)Poly::inv[n - l + 1] * g[l] % mod;
return ;
}
int mid = l + r >> 1;
solve(l,mid);
poly t1,t2,t3,t4,t;
int lim = 1,tot = mid - l + r - l + 2;
for(;lim < tot;lim <<= 1);
t1.resize(lim),t2.resize(lim),t3.resize(lim),t4.resize(lim),t.resize(lim);
for(register int i = 0;i < lim;++i)
t1.a[i] = t2.a[i] = t3.a[i] = t4.a[i] = t.a[i] = 0;
for(register int i = 0;i <= mid - l;++i)
t1.a[i] = (long long)(n - m) * g[i + l] % mod,
t3.a[i] = (long long)(i + l) * g[i + l] % mod;
for(register int i = 0;i <= r - l;++i)
t2.a[i] = i & 1 ? dec(0,Poly::ifac[i]) : Poly::ifac[i],
t4.a[i] = i & 1 ? Poly::ifac[i + 1] : dec(0,Poly::ifac[i + 1]);
if(mid - l <= 60 || r - l <= 100)
{
for(register int i = mid - l + 1;i <= r - l;++i)
for(register int j = 0;j <= i && j <= mid - l;++j)
t.a[i] = (t[i] + (long long)t1[j] * t2[i - j] + (long long)t3[j] * t4[i - j]) % mod;
}
else
{
t1.ntt(),t2.ntt(),t3.ntt(),t4.ntt();
for(register int i = 0;i < lim;++i)
t.a[i] = ((long long)t1[i] * t2[i] + (long long)t3[i] * t4[i]) % mod;
t.ntt(-1);
}
for(register int i = mid + 1;i <= r;++i)
g.a[i] = dec(g[i],t[i - l]);
solve(mid + 1,r);
}
poly t1,t2,a;
int main()
{
init();
scanf("%d%d%d",&n,&m,&k),n1 = k - 1,n2 = n - k;
for(register int i = 0;i < n - m;++i)
rest = (rest + (long long)(i & 1 ? mod - 1 : 1) * Poly::ifac[i] % mod * Poly::ifac[n - i] % mod * (fpow(n - m - i,n) - fpow(n - m - i - 1,n) + mod)) % mod;
rest = (long long)rest * Poly::fac[n] % mod;
g.resize(n),
g.a[0] = (long long)(m & 1 ? (mod - C(n + 1,m)) % mod : C(n + 1,m)) * (n - m + 1) % mod,
solve(0,n - 1);
b.resize(n);
for(register int i = 0;i < n;++i)
b.a[i] = Poly::ifac[i + 1];
b = b.inver(n);
p.resize(n),q.resize(n);
for(register int i = n2;i < n;++i)
p.a[n - 1 - i] = (long long)C(i,n2) * (n2 & 1 ? mod - 1 : 1) % mod * g[n - 1 - i] % mod;
for(register int i = n1;i < n;++i)
q.a[n - 1 - i] = (long long)C(i,n1) * (i - n1 & 1 ? mod - 1 : 1) % mod * g[n - 1 - i] % mod;
p = (p * b).modxn(n),q = (q * b).modxn(n),f.resize(n);
for(register int i = 0;i < n;++i)
f.a[i] = (long long)Poly::ifac[i] * (p[n - 1 - i] - q[n - 1 - i] + mod) % mod;
f = f.integ().modxn(n);
t1.resize(n),t2.resize(n);
for(register int i = 0;i < n;++i)
t1.a[n - 1 - i] = (long long)Poly::fac[i] * f[n - 1 - i] % mod,
t2.a[i] = Poly::ifac[i];
t1 *= t2,a.resize(n),c = 0;
for(register int i = 0;i < n;++i)
a.a[i] = (long long)Poly::ifac[n - 1 - i] * t1[i] % mod * Poly::fac[i] % mod * Poly::fac[n - 1 - i] % mod,
rest = dec(rest,a[i]),
c = (c + (long long)C(n - 1,i) * Poly::fac[i] % mod * Poly::fac[n - 1 - i]) % mod;
rest = (long long)rest * fpow(c,mod - 2) % mod;
for(register int i = 0;i < n;++i)
a.a[i] = (a[i] + (long long)C(n - 1,i) * Poly::fac[i] % mod * Poly::fac[n - 1 - i] % mod * rest) % mod;
for(register int i = 0;i < n;++i)
printf("%d%c",a[i]," \n"[i == n - 1]);
}