题目大意
给出(q)个查询，每次查询(n)个点的无根有标号仙人掌有多少个。
(qle 5 imes 10^4,n< 131072)
思路
因为这道题太难码了，所以先把题解写了再写代码（好奇怪啊）终于码出来了，果然还是( ext {vector})好用（雾
为了方便，我们下面的答案其实求的是有根有标号的答案，最后除以(n)就好了。我们设(c_n)为(n)个点时的答案，我们对其构造指数型生成函数(C(x)):
[C(x)=sum_{i=0} c_ifrac{x^i}{i!}
]
我们考虑钦定一个根，然后连边，显然要么连出去一条边，要么连出去一个环。需要注意的是，连边相当于把根与一个仙人掌相连，连环的环其实是仙人掌连成一个环。
于是，我们可以得到转移式:
[C(x)=xexp (C(x)+frac{1}{2}sum_{i=2}^{infty} C^i(x))
]
除以(2)是因为一个环无论正反都是一样，但是我们计算时却视作两种情况。
将上式用等比数列求和公式可以得到:
[C(x)=xexp(frac{2C(x)-C^2(x)}{2-2C(x)})
]
我们发现我们如果构造
[F(C(x))=xexp(frac{2C(x)-C^2(x)}{2-2C(x)})-C(x)
]
我们即是要求导(F(C(x)))的根，对此，我们可以使用多项式牛顿迭代法。我们就需要求到(F(C(x)))的导:
[F^{'}(C(x))=(xexp(frac{2C(x)-C^2(x)}{2-2C(x)})-C(x))^{'}
]
[=xexp (frac{2C(x)-C^2(x)}{2-2C(x)})(frac{2C(x)-C^2(x)}{2-2C(x)})^{'}-1
]
[=xexp (frac{2C(x)-C^2(x)}{2-2C(x)})frac{C^2(x)-2C(x)+2}{2C^2(x)-4C(x)+2}-1
]
于是，我们可以得到牛顿迭代法的式子:
[C(x)=C_0(x)-frac{F(C_0(x))}{F^{'}(C_0(x))}
]
[=C_0(x)-frac{xexp(frac{2C_0(x)-C_0^2(x)}{2-2C_0(x)})-C(x)}{xexp (frac{2C_0(x)-C_0^2(x)}{2-2C_0(x)})frac{C_0^2(x)-2C_0(x)+2}{2C_0^2(x)-4C_0(x)+2}-1}
]
[=C_0(x)-frac{2xexp (frac{2C_0(x)-C_0^2(x)}{2-2C_0(x)})-2C_0(x)}{(1+frac{1}{(C_0(x)-1)^2})xexp (frac{2C_0(x)-C_0^2(x)}{2-2C_0(x)})-2}
]
我们发现如果我们设(G=xexp (frac{2C_0(x)-C_0^2(x)}{2-2C_0(x)}))，那么式子就可以化为:
[C(x)=C_0(x)-frac{2G-2C_0(x)}{(1+frac{1}{(C_0(x)-1)^2})G-2}
]
( ext {Code})
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline", "no-stack-protector", "unroll-loops")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 1000005
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
typedef vector <int> poly;
int w[MAXN],rev[MAXN];
void init_ntt (){
int lim = 1 << 18;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
static unsigned long long d[MAXN];
for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
for (Int i = 1;i < lim;i <<= 1)
for (Int j = 0;j < lim;j += i << 1)
for (Int k = 0;k < i;++ k){
int x = d[i + j + k] * w[i + k] % mod;
d[i + j + k] = d[j + k] + mod - x,d[j + k] += x;
}
for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
if (type == -1){
reverse (a.begin() + 1,a.begin() + lim);
for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
}
#undef G
#undef Gi
}
poly operator + (poly a,poly b){
a.resize (max (SZ (a),SZ (b)));
for (Int i = 0;i < SZ (b);++ i) a[i] = add (a[i],b[i]);
return a;
}
poly operator - (poly a,poly b){
a.resize (max (SZ (a),SZ (b)));
for (Int i = 0;i < SZ (b);++ i) a[i] = dec (a[i],b[i]);
return a;
}
poly operator * (poly a,int b){
for (Int i = 0;i < SZ (a);++ i) a[i] = mul (a[i],b);
return a;
}
poly operator * (poly a,poly b){
int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
a.resize (lim),b.resize (lim);
ntt (a,lim,1),ntt (b,lim,1);
for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
ntt (a,lim,-1),a.resize (d);
return a;
}
poly operator << (poly a,int n){
a.resize (SZ (a) + n);
for (Int i = SZ (a) - 1;~i;-- i) a[i] = (i >= n ? a[i - n] : 0);
return a;
}
poly inv (poly a,int n){
poly b(1,inv (a[0])),c;
for (Int l = 4;(l >> 2) < n;l <<= 1){
c.resize (l >> 1);
for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
c.resize (l),b.resize (l);
ntt (c,l,1),ntt (b,l,1);
for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
ntt (b,l,-1),b.resize (l >> 1);
}
b.resize (n);
return b;
}
poly inv (poly a){return inv (a,SZ (a));}
poly der (poly a){
for (Int i = 0;i < SZ (a) - 1;++ i) a[i] = mul (a[i + 1],i + 1);
a.pop_back ();return a;
}
poly ine (poly a){
a.push_back (0);
for (Int i = SZ (a) - 1;i;-- i) a[i] = mul (a[i - 1],inv (i));
a[0] = 0;return a;
}
poly ln (poly a,int n){
a = ine (der (a) * inv (a));
a.resize (n);
return a;
}
poly ln (poly a){return ln (a,SZ (a));}
poly exp (poly a,int n){
poly b (1,1),c;
for (Int l = 2;(l >> 1) < n;l <<= 1){
b.resize (l),c = ln (b);
for (Int i = 0;i < l;++ i) c[i] = dec (i < n ? a[i] : 0,c[i]);
c[0] = add (c[0],1);
b = b * c,b.resize (l);
}
b.resize (n);
return b;
}
poly exp (poly a){return exp (a,SZ (a));}
int n,m,x;poly A;
poly work (int n){
poly a({0,1}),b,c;
for (Int l = 4;(l >> 1) < n;l <<= 1){
a.resize (l),c = inv (poly {1} - a),b = c * a,b.resize (l),b = (b + a) * inv (2);
b = exp (b),b = b << 1,c = c * c,c[0] = add (c[0],1),c.resize (l),c = c * inv (2) * b,c[0] = dec (c[0],1);
b = b - a;while (!b[0] && !c[0]) b.erase (b.begin()),c.erase (c.begin());
b.resize (l),c.resize (l);
a = a - b * inv (c),a.resize (l);
}
a.resize (n);
return a;
}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
int q,len,fac[MAXN],que[MAXN];
signed main(){
init_ntt(),read (q);for (Int i = 1;i <= q;++ i) read (que[i]),len = max (len,que[i]);
A = work (len);fac[0] = 1;for (Int i = 1;i <= len;++ i) fac[i] = mul (fac[i - 1],i);
for (Int i = 1;i <= q;++ i) write (1ll * A[que[i]] * fac[que[i] - 1] % mod),putchar ('
');
return 0;
}
话说预处理原根真的好快啊！！！