洛谷P5219 无聊的水题 I [prufer序列，生成函数，NTT]

传送门

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思路

有标号无根树的计数，还和度数有关，显然可以想到prufer序列。

问题就等价于求长度为\(n-2\)，值域为\([1,n]\)，出现次数最多的恰好出现\(m-1\)次，这样的序列有哪些。

恰好\(m-1\)次不好求，变成最多\(m-1\)减去最多\(m-2\)的方案数。

考虑指数型生成函数。设要求的最多为\(M\)，则设\(A(x)=\sum_{i=0}^M \frac{1}{i!}x^i\)，答案就为\((n-2)![x^{n-2}]A^n(x)\)，多项式快速幂即可。

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代码

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
	using namespace std;
	#define pii pair<int,int>
	#define fir first
	#define sec second
	#define MP make_pair
	#define rep(i,x,y) for (int i=(x);i<=(y);i++)
	#define drep(i,x,y) for (int i=(x);i>=(y);i--)
	#define go(x) for (int i=head[x];i;i=edge[i].nxt)
	#define templ template<typename T>
	#define sz 202020
	#define mod 998244353ll
	typedef long long ll;
	typedef double db;
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
	templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
	templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
	templ inline void read(T& t)
	{
		t=0;char f=0,ch=getchar();double d=0.1;
		while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
		while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
		if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
		t=(f?-t:t);
	}
	template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
	char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    inline void print(register int x)
    {
    	if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    	while(z[++Z]=x%10+48,x/=10);
    	while(sr[++C]=z[Z],--Z);sr[++C]='\n';
    }
	void file()
	{
		#ifndef ONLINE_JUDGE
		freopen("a.in","r",stdin);
		#endif
	}
	inline void chktime()
	{
		#ifndef ONLINE_JUDGE
		cout<<(clock()-t)/1000.0<<'\n';
		#endif
	}
	#ifdef mod
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
	ll inv(ll x){return ksm(x,mod-2);}
	#else
	ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
	#endif
//	inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int n,m;

ll fac[sz],_fac[sz];
void init(){fac[0]=_fac[0]=1;rep(i,1,sz-1) _fac[i]=inv(fac[i]=fac[i-1]*i%mod);}

namespace SOLVE
{
	int limit,r[sz];
	void NTT_init(int n)
	{
		int l=-1;limit=1;
		while (limit<=n+n) ++l,limit<<=1;
		rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
	}
	void NTT(ll *a,int type)
	{
		rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
		for (int mid=1;mid<limit;mid<<=1)
		{
			ll Wn=ksm(3,(mod-1)/mid>>1);if (type==-1) Wn=inv(Wn);
			for (int len=mid<<1,j=0;j<limit;j+=len)
			{
				ll w=1;
				for (int k=0;k<mid;k++,w=w*Wn%mod)
				{
					ll x=a[j+k],y=w*a[j+k+mid]%mod;
					a[j+k]=(x+y)%mod,a[j+k+mid]=(x-y+mod)%mod;
				}
			}
		}
		if (type==1) return;
		ll I=inv(limit);
		rep(i,0,limit-1) a[i]=a[i]*I%mod;
	}
	ll _a[sz],_b[sz];
	void mul(ll *a,ll *b,int n) // a*=b (mod x^n)
	{
		NTT_init(n);
		rep(i,0,n-1) _a[i]=a[i],_b[i]=b[i];
		rep(i,n,limit-1) _a[i]=_b[i]=0;
		NTT(_a,1);NTT(_b,1);
		rep(i,0,limit-1) _a[i]=_a[i]*_b[i]%mod;
		NTT(_a,-1);
		rep(i,0,n-1) a[i]=_a[i];
	}
	ll a[sz],b[sz];
	ll solve(int m)
	{
		rep(i,0,n) a[i]=b[i]=0;
		rep(i,0,m-1) a[i]=_fac[i];
		b[0]=1;
		int t=n;
		for (;t;t>>=1,mul(a,a,n)) if (t&1) mul(b,a,n);
		return fac[n-2]*b[n-2]%mod;
	}
}
using SOLVE::solve;

int main()
{
	file();
	read(n,m);
	init();
	cout<<(solve(m)-solve(m-1)+mod)%mod;
}