Luogu P5349 幂

大力数学题,发现自己好久没写多项式水平急速下降,求逆都要写挂233

首先看到关于多项式的等比数列求和,我们容易想到先求出每一项的系数然后最后累加起来即可,即设\(f_i=\sum_{n=0}^{\infty} n^kr^n\),那么最后\(ans=\sum_{i=0}^m a_if_i\)

然后你直接去OEIS就找到了它的生成函数,一波搞定233

好了我们认真开始推式子,\(f_i=\sum_{n=0}^{\infty} n^kr^n\)

\(r\cdot f_i=\sum_{n=1}^\infty (n-1)^kr^n\)

\((1-r)f_i=\sum_{n=1}^\infty (n^k-(n-1)^k)r^n=r\cdot \sum_{n=0}^\infty ((n+1)^k-n^k)r^n\)

上式等于\(r\cdot \sum_{n=0}^\infty\sum_{j=0}^{i-1} C_i^j\cdot n^i r^n=r\cdot \sum_{j=0}^{i-1} C_i^j\cdot \sum_{n=0}^\infty n^i r^n=r\cdot\sum_{j=0}^{i-1} C_i^j\cdot f_j\)

即有\(f_i=\frac{r}{1-r}\cdot \sum_{j=0}^{i-1} C_i^j\cdot f_j\)

把组合数展开就有\(\frac{f_i}{i!}=\sum_{j=0}^{i-1} \frac{f_j}{j!}(\frac{r}{1-r}\cdot\frac{1}{(k-i)!})\)

看出来这是个分治NTT的形式,用类似于板子题的方法转化为多项式求逆解决

具体地,令\(F(x)=1-\frac{1}{i!}\cdot\frac{r}{1-r}\),对\(F(x)\)求逆得出\(G(x)\)再乘回\(i!\)即可求出\(f(x)\)

CODE

#include<cstdio>
#include<cctype>
#include<algorithm>
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
const int N=100005,mod=998244353;
int n,r,F[N<<3],G[N<<3],fact[N],inv[N],A[N],invr,mtir,ans;
class FileInputOutput
{
    private:
        static const int S=1<<21;
        #define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
        char Fin[S],*A,*B;
    public:
        Tp inline void read(T& x)
        {
            x=0; char ch; while (!isdigit(ch=tc()));
            while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
        }
        #undef tc
}File;
inline int sum(CI x,CI y)
{
    int t=x+y; return t>=mod?t-mod:t;
}
inline int sub(CI x,CI y)
{
    int t=x-y; return t<0?t+mod:t;
}
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
    for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline void init(CI n)
{
    RI i; for (fact[0]=i=1;i<=n;++i) fact[i]=1LL*fact[i-1]*i%mod;
    for (inv[n]=quick_pow(fact[n]),i=n-1;~i;--i) inv[i]=1LL*inv[i+1]*(i+1)%mod;
}
class Poly_Solver
{
    private:
        int rev[N<<3],T[N<<3],lim,p;
        inline void init(CI n)
        {
            for (lim=1,p=0;lim<=(n<<1);lim<<=1,++p);
            for (RI i=0;i<lim;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<p-1);
        }
        inline void swap(int& x,int& y)
        {
            int t=x; x=y; y=t;
        }
        inline void NTT(int *f,CI opt)
        {
            RI i; for (i=0;i<lim;++i) if (i<rev[i]) swap(f[i],f[rev[i]]);
            for (i=1;i<lim;i<<=1)
            {
                int m=i<<1,D=quick_pow(3,~opt?(mod-1)/m:mod-1-(mod-1)/m);
                for (RI j=0;j<lim;j+=m)
                {
                    int W=1; for (RI k=0;k<i;++k,W=1LL*W*D%mod)
                    {
                        int x=f[j+k],y=1LL*f[i+j+k]*W%mod;
                        f[j+k]=sum(x,y); f[i+j+k]=sub(x,y);
                    }
                }
            }
            if (!~opt)
            {
                int Inv=quick_pow(lim); for (i=0;i<lim;++i) f[i]=1LL*f[i]*Inv%mod;
            }
        }
        inline void inv(int *F,int *G,CI n)
        {
            if (n==1) return (void)(G[0]=quick_pow(F[0]));
            inv(F,G,n+1>>1); init(n<<1); copy(F,F+n,T);
            NTT(T,1); NTT(G,1); for (RI i=0;i<lim;++i)
            G[i]=1LL*sub(2,1LL*T[i]*G[i]%mod)*G[i]%mod;
            NTT(G,-1); fill(T+n,T+lim,0); fill(G+n,G+lim,0);
        }
    public:
        inline void get_inv(int *F,int *G,CI n)
        {
            for (lim=1;lim<=(n<<1);lim<<=1);
            fill(T,T+lim,0); fill(G,G+lim,0); inv(F,G,n);
        }
}P;
int main()
{
    //freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
    File.read(n); File.read(r); invr=quick_pow(sub(1,r)); mtir=1LL*invr*r%mod;
    RI i; for (init(n),i=0;i<=n;++i) File.read(A[i]),F[i]=sum(mod,-1LL*inv[i]*mtir%mod);
    for (F[0]=1,P.get_inv(F,G,n+1),i=0;i<=n;++i) G[i]=1LL*G[i]*fact[i]%mod*invr%mod;
    for (i=0;i<=n;++i) ans=sum(ans,1LL*G[i]*A[i]%mod); return printf("%d",ans),0;
}
posted @ 2019-05-08 21:43 hl666 阅读(...) 评论(...) 编辑 收藏