Codeforces Round 641 F2. Slime and Sequences （生成函数）（拉格朗日反演）
传送门
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注意到好的序列到一个排列是一个双射，在序列上的数 a p i a_{p_i} api 等于 i i i 前方的小于号个数加一，其中小于号指的是相邻两个数的大小关系，下面我们对这个排列进行计数
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f i , j f_{i,j} fi,j 表示长度为 i i i 至少 j j j 个 < < < 的个数，恰为斯特林数 S i , i − j S_{i,i-j} Si,i−j，每一组内钦定递增
g i , j g_{i,j} gi,j 表示正好 j j j 个 < < < 的个数（可以 O ( n 2 ) O(n^2) O(n2) 简单 d p dp dp）
g i , j = ∑ k ≥ j ( − 1 ) k − j ( k j ) f i , j g_{i,j}=\sum_{k\ge j}(-1)^{k-j}\binom{k}{j}f_{i,j} gi,j=k≥j∑(−1)k−j(jk)fi,j
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考虑每个数的贡献，钦定它前面有 i i i 个 < < <
a n s i = ∑ j = i n g j , i ∗ ( n j ) ( n − j ) ! = ∑ j = i n ∑ k = i n S j , j − k ( − 1 ) k − i ( k i ) n ! j ! = n ! ∑ k = i n ( − 1 ) k − i ( k i ) ∑ j = k n [ x j ] ( e x − 1 ) j − k ans_i=\sum_{j=i}^ng_{j,i}*\binom{n}{j}(n-j)!\\=\sum_{j=i}^n\sum_{k=i}^nS_{j,j-k}(-1)^{k-i}\binom{k}{i}\frac{n!}{j!}\\=n!\sum_{k=i}^n(-1)^{k-i}\binom{k}{i}\sum_{j=k}^n[x^j](e^x-1)^{j-k} ansi=j=i∑ngj,i∗(jn)(n−j)!=j=i∑nk=i∑nSj,j−k(−1)k−i(ik)j!n!=n!k=i∑n(−1)k−i(ik)j=k∑n[xj](ex−1)j−k -
考虑快速求出 f k = ∑ j = k n [ x j ] ( e x − 1 ) j − k = [ x k ] ∑ j = 1 n − k ( e x − 1 x ) j = [ x k ] F n − k + 1 − F F − 1 f_k=\sum_{j=k}^n[x^j](e^x-1)^{j-k}\\=[x^k]\sum_{j=1}^{n-k}(\frac{e^x-1}{x})^{j}\\=[x^k]\frac{F^{n-k+1}-F}{F-1} fk=j=k∑n[xj](ex−1)j−k=[xk]j=1∑n−k(xex−1)j=[xk]F−1Fn−k+1−F
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考虑对每个 k k k 求出（ F = e x − 1 x F=\frac{e^x-1}{x} F=xex−1 ， G = e x − 1 G=e^x-1 G=ex−1）
[ x k ] F n − k + 1 1 − F = [ x n + 1 ] G n − k + 1 1 − F = [ x n + 1 y n − k + 1 ] ∑ i ≥ 0 ( y G ) i 1 − F = [ x n + 1 y n − k + 1 ] 1 1 − y G 1 1 − F [x^k]\frac{F^{n-k+1}}{1-F}=[x^{n+1}]\frac{G^{n-k+1}}{1-F}\\=[x^{n+1}y^{n-k+1}]\sum_{i\ge 0}\frac{(yG)^i}{1-F}\\=[x^{n+1}y^{n-k+1}]\frac{1}{1-yG}\frac{1}{1-F} [xk]1−FFn−k+1=[xn+1]1−FGn−k+1=[xn+1yn−k+1]i≥0∑1−F(yG)i=[xn+1yn−k+1]1−yG11−F1
构造 ϕ ( x ) \phi(x) ϕ(x) 使得 F = ϕ ( G ) F=\phi(G) F=ϕ(G)，则 G = x F , G ϕ ( G ) = x G=xF,\frac{G}{\phi(G)}=x G=xF,ϕ(G)G=x
聪明的你们肯定看出构造的用意了，令 T ( x ) = x ϕ ( x ) T(x)=\frac{x}{\phi(x)} T(x)=ϕ(x)x，拉个朗日反演
令 H ( x ) = H ( T ( G ) ) = 1 1 − y x 1 1 − ϕ ( x ) H(x)=H(T(G))=\frac{1}{1-yx}\frac{1}{1-\phi(x)} H(x)=H(T(G))=1−yx11−ϕ(x)1
[ x n + 1 ] [ y n − k + 1 ] H ( G ) = [ x n ] [ y n − k + 1 ] 1 n + 1 d H d x ( x T ) n + 1 = 1 n + 1 [ y n − k + 1 ] [ x n ] ( − ( ( 1 − x y ) ( 1 − ϕ ( x ) ) ) ′ ( 1 − x y ) 2 ( 1 − ϕ ( x ) ) 2 ϕ ( x ) n + 1 ) = 1 n + 1 [ y n − k + 1 ] [ x n ] ( y ( 1 − ϕ ( x ) ) + ϕ ( x ) ′ ( 1 − x y ) ( 1 − x y ) 2 ( 1 − ϕ ( x ) ) 2 ) ϕ ( x ) n + 1 = 1 n + 1 [ y n − k + 1 ] [ x n ] ( y ( 1 − x y ) 2 ( 1 − ϕ ( x ) ) + ϕ ( x ) ′ ( 1 − x y ) ( 1 − ϕ ( x ) ) 2 ϕ ( x ) n + 1 ) [x^{n+1}][y^{n-k+1}]H(G)=[x^n][y^{n-k+1}]\frac{1}{n+1}\frac{\text{d}H}{\text{d}x}(\frac{x}{T})^{n+1}\\=\frac{1}{n+1}[y^{n-k+1}][x^n](\frac{-((1-xy)(1-\phi(x)))'}{(1-xy)^2(1-\phi(x))^2}\phi(x)^{n+1})\\=\frac{1}{n+1}[y^{n-k+1}][x^n](\frac{y(1-\phi(x))+\phi(x)'(1-xy)}{(1-xy)^2(1-\phi(x))^2})\phi(x)^{n+1}\\ =\frac{1}{n+1}[y^{n-k+1}][x^n](\frac{y}{(1-xy)^2(1-\phi(x)) }+\frac{\phi(x)'}{(1-xy)(1-\phi(x))^2}\phi(x)^{n+1}) [xn+1][yn−k+1]H(G)=[xn][yn−k+1]n+11dxdH(Tx)n+1=n+11[yn−k+1][xn]((1−xy)2(1−ϕ(x))2−((1−xy)(1−ϕ(x)))′ϕ(x)n+1)=n+11[yn−k+1][xn]((1−xy)2(1−ϕ(x))2y(1−ϕ(x))+ϕ(x)′(1−xy))ϕ(x)n+1=n+11[yn−k+1][xn]((1−xy)2(1−ϕ(x))y+(1−xy)(1−ϕ(x))2ϕ(x)′ϕ(x)n+1)
= 1 n + 1 [ y n − k + 1 ] [ x n ] ϕ ( x ) n + 1 ( 1 ( 1 − ϕ ( x ) ) ∑ t ≥ 0 x t y t + 1 ( t + 1 ) + ϕ ( x ) ′ ( 1 − ϕ ( x ) ) 2 ∑ t ≥ 0 x t y t ) = 1 n + 1 [ x n ] ϕ ( x ) n + 1 { 1 ( 1 − ϕ ( x ) ) ( n − k + 1 ) x n − k + ϕ ( x ) ′ ( 1 − ϕ ( x ) ) 2 x n − k + 1 } = 1 n + 1 ( [ x k ] ( n − k + 1 ) ϕ ( x ) n + 1 1 − ϕ ( x ) + [ x k − 1 ] ϕ ( x ) ′ ϕ ( x ) n + 1 ( 1 − ϕ ( x ) ) 2 ) =\frac{1}{n+1}[y^{n-k+1}][x^n]\phi(x)^{n+1}(\frac{1}{(1-\phi(x))}\sum_{t\ge 0}x^ty^{t+1}(t+1)+\frac{\phi(x)'}{(1-\phi(x))^2}\sum_{t\ge 0}x^ty^t)\\=\frac{1}{n+1}[x^n]\phi(x)^{n+1}\{\frac{1}{(1-\phi(x))}(n-k+1)x^{n-k}+\frac{\phi(x)'}{(1-\phi(x))^2}x^{n-k+1}\}\\=\frac{1}{n+1}([x^k]\frac{(n-k+1)\phi(x)^{n+1}}{1-\phi(x)}+[x^{k-1}]\frac{\phi(x)'\phi(x)^{n+1}}{(1-\phi(x))^2}) =n+11[yn−k+1][xn]ϕ(x)n+1((1−ϕ(x))1t≥0∑xtyt+1(t+1)+(1−ϕ(x))2ϕ(x)′t≥0∑xtyt)=n+11[xn]ϕ(x)n+1{(1−ϕ(x))1(n−k+1)xn−k+(1−ϕ(x))2ϕ(x)′xn−k+1}=n+11([xk]1−ϕ(x)(n−k+1)ϕ(x)n+1+[xk−1](1−ϕ(x))2ϕ(x)′ϕ(x)n+1)
- ϕ ( x ) = x ln ( x + 1 ) \phi(x)=\frac{x}{\ln(x+1)} ϕ(x)=ln(x+1)x， O ( n log n ) O(n\log n) O(nlogn)
#include
#define pb push_back
#define cs const
#define poly vector
using namespace std;
cs int Mod = 998244353;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
int ksm(int a, int b){ int as=1; for(;b;b>>=1,a=mul(a,a)) if(b&1) as=mul(as,a); return as; }
void Add(int &a, int b){ a = add(a, b); }
void Dec(int &a, int b){ a = dec(a, b); }
void Mul(int &a, int b){ a = mul(a, b); }
cs int K = 18, N = 1 << K | 5;
int n, fc[N], ifc[N], iv[N];
int *w[K+1], rv[N], bit, up;
void output(poly a){ for(int i=0; isize(); i++) cout<" "; puts(""); }
void fc_init(int n){
fc[0]=fc[1]=ifc[0]=ifc[1]=1;
for(int i=2; i<=n; i++) fc[i]=mul(fc[i-1],i);
for(int i=2; i<=n; i++) ifc[i]=mul(ifc[i-1],iv[i]);
}
void NTT_init(){
for(int i=1; i<=K; i++) w[i]=new int[1<<(i-1)];
int wn=ksm(3,(Mod-1)/(1<)); w[K][0]=1;
for(int i=1; i<(1<<(K-1)); i++) w[K][i]=mul(w[K][i-1],wn);
for(int i=K-1;i;i--) for(int j=0;j<(1<<(i-1));j++) w[i][j]=w[i+1][j<<1];
iv[0]=iv[1]=1; for(int i=2; i<=(1<); i++) iv[i]=mul(Mod-Mod/i,iv[Mod%i]);
}
void init(int deg){
up=1; bit=0; while(up) up<<=1,++bit;
for(int i=0; i; i++) rv[i]=(rv[i>>1]>>1)|((i&1)<<(bit-1));
}
poly operator - (poly a, poly b){
if(a.size()size()) a.resize(b.size());
for(int i=0; i<(int)b.size(); i++) Dec(a[i],b[i]); return a;
}
poly divx(poly a){
for(int i=0; i<(int)a.size(); i++) a[i]=a[i+1];
return a.pop_back(), a;
}
void NTT(poly &a, int typ=1){
for(int i=0; i; i++) if(i) swap(a[i],a[rv[i]]);
for(int i=1,l=1;i;i<<=1,++l)
for(int j=0; j; j+=(i<<1))
for(int k=0; k; k++){
int x=a[k+j], y=mul(w[l][k],a[k+j+i]);
a[k+j]=add(x,y); a[k+j+i]=dec(x,y);
} if(typ==-1){
reverse(a.begin()+1,a.end());
for(int i=0,iv=ksm(up,Mod-2); i; i++) Mul(a[i],iv);
}
}
poly operator * (poly a, poly b){
int dg=a.size()+b.size()-1;
if(a.size()<=32||b.size()<=32){
poly c(dg,0);
for(int i=0; i<(int)a.size(); i++)
for(int j=0; j<(int)b.size(); j++)
Add(c[i+j],mul(a[i],b[j])); return c;
} init(dg); a.resize(up); b.resize(up); NTT(a); NTT(b);
for(int i=0; i; i++) Mul(a[i],b[i]);
NTT(a,-1); a.resize(dg); return a;
}
poly inv(poly a, int deg){
poly b(1,ksm(a[0],Mod-2)),c;
for(int lim=2; (lim>>1); lim<<=1){
c.resize(lim); init(lim<<1);
for(int i=0; i; i++) c[i]=i<(int)a.size()?a[i]:0;
c.resize(up); b.resize(up); NTT(c); NTT(b);
for(int i=0; i; i++) Mul(b[i],dec(2,mul(b[i],c[i])));
NTT(b,-1); b.resize(lim);
} b.resize(deg); return b;
}
poly deriv(poly a){
for(int i=0; i+1<(int)a.size(); i++) a[i]=mul(i+1,a[i+1]);
a.pop_back(); return a;
}
poly integ(poly a){
a.pb(0);
for(int i=a.size()-1;i;i--) a[i]=mul(iv[i],a[i-1]);
a[0]=0; return a;
}
poly ln(poly a, int dg){
if(dg==-1) dg=a.size();
a=integ(deriv(a)*inv(a,dg)); a.resize(dg); return a;
}
poly Exp(poly a, int dg){
if(dg==-1) dg = a.size(); poly b(1,1), c;
for(int lim=2; lim<(dg<<1); lim<<=1){
c=ln(b,lim); Dec(c[0],1);
for(int i=0; i; i++) c[i]=dec(i<(int)a.size()?a[i]:0,c[i]);
b=b*c; b.resize(lim);
} b.resize(dg); return b;
}
poly polypw(poly a, int k, int dg){
a = ln(a,dg);
for(int i=0; i<(int)a.size(); i++) Mul(a[i],k);
return Exp(a,dg);
}
int main(){
#ifdef FSYolanda
freopen("1.in","r",stdin);
#endif
scanf("%d",&n); NTT_init(); fc_init(n+5);
poly phi(2,1); phi = ln(phi,n+3); phi = inv(divx(phi),n+3);
poly t = poly(1,1) - phi;
t = inv(divx(t),n+3); poly tt = t * t; tt.resize(n+3);
poly mt = polypw(phi,n+1,n+3), dt = deriv(phi);
t = mt * t; tt = mt * tt; t.resize(n+3); tt.resize(n+3);
tt = tt * dt; tt.resize(n+3);
poly em(n+3,0);
for(int i=0; i<=n+2; i++) em[i]=ifc[i+1];
em=em*inv(divx(poly(1,1)-em),n+3);
poly f(n+1,0);
for(int i=0,iv=ksm(n+1,Mod-2); i; i++){
f[i] = mul(iv,add(mul(t[i+1],n-i+1), tt[i+1]));
f[i] = dec(em[i+1],f[i]); Mul(f[i],fc[i]);
} poly g(n,0);
for(int i=0; i; i++) g[i]=(i&1)?dec(0,ifc[i]):ifc[i];
reverse(g.begin(),g.end()); g = g * f;
for(int i=0; i; i++) cout<<mul(ifc[i],mul(fc[n],g[i+n-1]))<<" ";
return 0;
}