洛谷P5219 无聊的水题 I [prufer序列，生成函数，NTT]

传送门

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。

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;
}

洛谷P5219 无聊的水题 I [prufer序列，生成函数，NTT]

原文：https://www.cnblogs.com/p-b-p-b/p/10440527.html