[luogu P5349] 幂 解题报告 (分治FFT)

interlinkage:

https://www.luogu.org/problemnew/show/P5349

description:

solution:

设$g(x)=\sum_{n=0}^{∞}n^xr^n$

$rg(x)=\sum_{n=0}^{∞}n^xr^{n+1}=\sum_{n=1}^{∞}(n-1)^xr^n$

$g(x)=\sum_{n=1}^{∞}n^xr^n(x>0)$(注意$x>0$这个条件，$x=0$的时候这个不符合)

$(1-r)g(x)=\sum_{n=1}^{∞}(n^x-(n-1)^x)r^n=r\sum_{n=0}^{∞}r^n((n+1)^x-n^x)=r\sum_{n=0}^{∞}r^n\sum_{j=0}^{x-1}\dbinom{x}{j}n^j$

$=r\sum_{j=0}^{x-1}\dbinom{x}{j}\sum_{n=0}^{∞}n^jr^n=r\sum_{j=0}^{x-1}\dbinom{x}{j}g(j)$

于是$g(x)=\frac{r}{1-r}\sum_{j=0}^{x-1}\dbinom{x}{j}g(j)$

继续化简得到$\frac{g(x)}{x!}=\sum_{j=0}^{x-1}\frac{g(j)}{j!}(\frac{r}{1-r}*\frac{1}{(x-j)!})$

这个显然可以用分治$FFT$来做

值得注意的是$g(0)=\frac{1}{1-r}$，而不是$\frac{r}{1-r}$，因为在这里$0^0$的值实际上是算$1$的

直接分治的话复杂度为$O(nlognlogn)$，多项式求逆时间复杂度为$O(nlogn)$

code:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;

const int N=4e5+;
const ll mo=;
int m;
ll r;
ll a[N],wn[N],R[N],fac[N],inv[N];
inline ll read()
{
    char ch=getchar();ll s=,f=;
    while (ch<''||ch>'') {if (ch=='-') f=-;ch=getchar();}
    while (ch>=''&&ch<='') {s=(s<<)+(s<<)+ch-'';ch=getchar();}
    return s*f;
}
ll qpow(ll a,ll b)
{
    ll re=;
    for (;b;b>>=,a=a*a%mo) if (b&) re=re*a%mo;
    return re;
}
void pre()
{
    for (int i=;i<=;i++)
    {
        ll t=1ll<<i;
        wn[i]=qpow(,(mo-)/t);
    }
}
void ntt(int limit,ll *a,int type)
{
    for (int i=;i<limit;i++) if (i<R[i]) swap(a[i],a[R[i]]);
    for (int len=,id=;len<limit;len<<=)
    {
        ++id;
        for (int k=;k<limit;k+=(len<<))
        {
            ll w=;
            for (int l=;l<len;l++,w=w*wn[id]%mo)
            {
                ll Nx=a[k+l],Ny=w*a[k+len+l]%mo;
                a[k+l]=(Nx+Ny)%mo;
                a[k+len+l]=((Nx-Ny)%mo+mo)%mo;
            }
        }
    }
    if (type==) return;
    for (int i=;i<limit/;i++) swap(a[i],a[limit-i]);
    ll inv=qpow(limit,mo-);
    for (int i=;i<limit;i++) a[i]=a[i]%mo*inv%mo;
}
ll A[N],B[N];
void cdqfft(ll *a,ll *b,int l,int r)
{
    if (l==r) return;
    int mid=l+r>>;
    cdqfft(a,b,l,mid);

    int limit=,L=;
    while (limit<=(r-l+)*) limit<<=,++L;
    for (int i=;i<=limit;i++) R[i]=(R[i>>]>>)|((i&)<<(L-));

    for (int j=;j<=limit;j++) A[j]=,B[j]=;
    for (int j=l;j<=mid;j++) A[j-l]=a[j];
    for (int j=;j<=r-l;j++) B[j]=b[j];
    ntt(limit,A,);ntt(limit,B,);
    for (int i=;i<=limit;i++) A[i]=A[i]*B[i]%mo;
    ntt(limit,A,-);
    for (int x=mid+;x<=r;x++) a[x]=(a[x]+A[x-l])%mo;
    cdqfft(a,b,mid+,r);
}
ll g[N],f[N];
int main()
{
    pre();
    m=read();r=read();
    for (int i=;i<=m;i++) a[i]=read();
    fac[]=inv[]=;
    for (int i=;i<=m;i++) fac[i]=fac[i-]*i%mo;
    inv[m]=qpow(fac[m],mo-);
    for (int i=m-;i>=;i--) inv[i]=inv[i+]*(i+)%mo;
    f[]=qpow(-r+mo,mo-)%mo;
    for (int i=;i<=m;i++) g[i]=inv[i]*f[]%mo*r%mo;
    cdqfft(f,g,,m);
    ll ans=;
    for (int i=;i<=m;i++) ans=(ans+a[i]*f[i]%mo*fac[i]%mo)%mo;
    printf("%lld\n",ans);
    return ;
}