Loj #6703 -「清华集训 2017」生成树计数

标签：static   push   inline   alc   isp   def   add   +=   cto   

\[ \sum _{\sum v_i = n-2} \prod (a_i ^{v_i+1} * (v_i+1) ^ m /v_i!) *(\sum (v_i+1)^m) \]

将 \(\sum (v_i+1)^m\) 中的贡献分开算。

我们有两个生成函数。

第一个：
\[ \sum _i a^{i+1} * (v_i+1)^m x^i / i! \]
第二个：
\[ \sum _i a^{i+1} * (i+1)^{2m} x^i / i! \]
先将 \(\prod a\) 算到前面。

令：
\[ A(x) =\sum x^i * (i+1)^{m} /i! \B(x) =\sum x^i * (i+1)^{2m} /i! \]
我们要算的是：
\[ \sum _i B(a_ix)/A(a_ix) * \prod A(a_jx) \]
后面可以ln，exp。

然后发现泥算个等幂和就行了。

补充一下等幂和怎么算：
生成函数是：
\[ \sum \frac 1 {1-a_ix} \]
因为
\[ \ln' (f(x)) = \frac {f'(x)}{f(x)} \\ln' (1-a_ix) = \frac {-a_i} {1-a_ix} = -a_i - a_i^2 x \]
所以可以一个分治FFT算。

最终复杂度$n\log ^2n $

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+5;
typedef long long ll;

const int mod=998244353;
int add(int a,int b){a+=b;return a>=mod?a-mod:a;}
int sub(int a,int b){a-=b;return a<0?a+mod:a;}
int mul(int a,int b){return (ll)a*b%mod;}
int qpow(int a,int b){int ret=1;for(;b;b>>=1,a=mul(a,a))if(b&1)ret=mul(ret,a);return ret;}
/*math*/

namespace Template_Poly{
    typedef vector<int> Poly;
    int rev[N];
    Poly Poly_add(Poly A,Poly B){
        A.resize(max(A.size(),B.size()));
        for(size_t i=0;i<B.size();i++)A[i]=add(A[i],B[i]);
        return A;
    }
    Poly Poly_sub(Poly A,Poly B){
        A.resize(max(A.size(),B.size()));
        for(size_t i=0;i<B.size();i++)A[i]=sub(A[i],B[i]);
        return A;
    }
    void DFT(int *t,int n,int type){
        int l=0;while(1<<l<n)++l;
        for(int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
        for(int i=0;i<n;i++)if(rev[i]>i)swap(t[rev[i]],t[i]);
        for(int step=1;step<n;step<<=1){
            int wn=qpow(3,(mod-1)/(step<<1));
            for(int i=0;i<n;i+=step<<1){
                int w=1;
                for(int k=0;k<step;k++,w=mul(w,wn)){
                    int x=t[i+k],y=mul(t[i+k+step],w);
                    t[i+k]=add(x,y),t[i+k+step]=sub(x,y);
                }
            }
        }
        if(type==1)return;
        for(int i=1;i<n-i;i++)swap(t[i],t[n-i]);
        int inv=qpow(n,mod-2);
        for(int i=0;i<n;i++)t[i]=mul(t[i],inv);
    }
    Poly NTT(Poly A,int n,Poly B,int m){
        static Poly res,PolA,PolB;
        PolA=A,PolB=B;
        int len=1;while(len < n+m)len<<=1;
        res.resize(len);
        PolA.resize(len),PolB.resize(len);
        DFT(&PolA[0],len,1);DFT(&PolB[0],len,1);
        for(int i=0;i<len;i++) res[i]= mul(PolA[i],PolB[i]);
        DFT(&res[0],len,-1);
        res.resize(n+m-1);
        return res;
    }
    Poly NTT(Poly A,Poly B){
        return NTT(A,A.size(),B,B.size());
    }
    Poly Poly_inv(Poly A,int n){
        if(n==1)return Poly(1,qpow(A[0],mod-2));
        int len=1<<((int)ceil(log2(n))+1);
        Poly x=Poly_inv(A,(n+1)>>1),y;
        x.resize(len),y.resize(len);
        for(int i=0;i<n;i++)y[i]=A[i];
        DFT(&x[0],len,1),DFT(&y[0],len,1);
        for(int i=0;i<len;i++)x[i]=mul(x[i],sub(2,mul(x[i],y[i])));
        DFT(&x[0],len,-1);
        x.resize(n);
        return x;
    }
    Poly Poly_inv(Poly A){
        return Poly_inv(A,A.size());
    }
    Poly Deri(Poly A){
        int n=A.size();
        for(int i=1;i<n;i++)A[i-1]=mul(A[i],i);
        A.resize(n-1);
        return A;
    }

    Poly Inte(Poly A){
        int n=A.size();
        A.resize(n+1);
        for(int i=n;i;i--)A[i]=mul(A[i-1],qpow(i,mod-2));
        A[0]=0;
        return A;
    }

    Poly ln(Poly A){
        int len=A.size();
        A=Inte(NTT(Deri(A),Poly_inv(A)));
        A.resize(len);
        return A;
    }

    Poly exp(Poly A,int n){
        if(n==1)return Poly(1,1);
        Poly x=exp(A,(n+1)>>1),y;
        x.resize(n);
        y=ln(x);
        for(int i=0;i<n;i++)y[i]=sub(A[i],y[i]);
        y[0]++;
        x=NTT(x,y);
        x.resize(n);
        return x;
    }
    Poly exp(Poly A){
        return exp(A,A.size());
    }

    Poly sqrt(Poly A,int n){
        if(n==1)return Poly(1,1);
        Poly x=sqrt(A,(n+1)>>1),y;
        x.resize(n),y.resize(n);
        for(int i=0;i<n;i++)y[i]=A[i];
        x=Poly_add(NTT(Poly_inv(x),y),x);
        int inv2=qpow(2,mod-2);
        for(int i=0;i<n;i++)
            x[i]=mul(x[i],inv2);
        x.resize(n);
        return x;
    }
    Poly sqrt(Poly A){
        return sqrt(A,A.size());
    }
    Poly rever(Poly A){
        reverse(A.begin(),A.end());
        return A;
    }
    void div(Poly A,Poly B,Poly &C,Poly &D){
        int n=A.size(),m=B.size();
        Poly ra=rever(A),rb=rever(B);
        ra.resize(n-m+1),rb.resize(n-m+1);
        C=NTT(ra,Poly_inv(rb));
        C.resize(n-m+1);
        C=rever(C);
        D=Poly_sub(A,NTT(B,C));
        D.resize(m);
    }
}
using namespace Template_Poly;
typedef Poly poly;
int n,m;
int a[N],fac[N],ifac[N];
inline void init(int n = 100010){
    fac[0]=ifac[0]=1;for(int i=1;i<=n;i++)fac[i] = mul(fac[i-1],i);
    ifac[n]=qpow(fac[n],mod-2);for(int i=n-1;i;i--)ifac[i] = mul(ifac[i+1],i+1);
}

poly solve(int l,int r){
    if(l==r){
        poly ret(2,1);
        ret[1]=sub(0,a[l]);
        return ret;
    }
    int mid=(l+r)>>1;
    return NTT(solve(l,mid),solve(mid+1,r));
}
poly func;
inline void calc(){
    func = solve(1,n);
    func = ln(func);
    // for(int i=0;i<func.size();i++)cout << func[i] << " ";puts("");
    func = Deri(func);
    func.push_back(0);
    for(int i=(int)func.size()-1;i;i--)func[i] = sub(0,func[i-1]);
    func[0]=n;
    // for(int i=0;i<func.size();i++)cout << func[i] << " ";puts("");
}
poly A,B,C;
poly ret;
void Doit(){
    A.resize(n-1),B.resize(n-1);
    for(int i=0;i<=n-2;i++){
        A[i] = mul(qpow(i+1, m),ifac[i]);
        B[i] = mul(qpow(i+1, 2*m),ifac[i]);
    }
    //A[0] = 1
    // B = Poly_inv(B);
    C = NTT(B,Poly_inv(A));
    C.resize(n-1);
    A = ln(A);
    for(int i=0;i<=n-2;i++){
        A[i] = mul(A[i], func[i]);
        C[i] = mul(C[i], func[i]);
    }
    A = exp(A);
    ret = NTT(A,C);
    int ans=ret[n-2];
    for(int i=1;i<=n;i++)ans=mul(ans,a[i]);
    ans = mul(ans, fac[n-2]);
    printf("%d\n",ans);
}

int main()
{
    init();
    cin >> n >> m;
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    calc();
    Doit();
}

Loj #6703 -「清华集训 2017」生成树计数

标签：static   push   inline   alc   isp   def   add   +=   cto   

原文地址：https://www.cnblogs.com/weiyanpeng/p/12028247.html