UOJ269. 【清华集训2016】如何优雅地求和 [生成函数]

传送门

思路

神仙题.jpg

脑子一抽，想把\(f(x)\)表示成下降幂的形式，也就是

\[f(x)=\sum_{i=0}^m f_ix_{(i)}\\
x_{(i)}=\prod_{k=0}^{i-1}(x-k)=[x\ge i]\frac{x!}{(x-i)!}
\]

这样有什么好处呢？回到原来的式子，我们有

\[\begin{align*}
&\sum_{k=0}^n f(k){n\choose k}x^k(1-x)^{n-k}\\
=&\sum_{k=0}^n \sum_{i=0}^m f_ik_{(i)}{n\choose k}x^k(1-x)^{n-k}\\
=&\sum_{i=0}^mf_i \sum_{k=i}^n \frac{n!}{(n-k)!(k-i)!} x^k(1-x)^{n-k}\\
=&\sum_{i=0}^m f_i n_{(i)}\sum_{k=i}^n {n-i\choose k-i} x^{k}(1-x)^{n-k}\\
=&\sum_{i=0}^m f_in_{(i)}x^i
\end{align*}
\]

可以非常快地求出来。

还有一个问题：\(f_i\)怎么求？

再一次脑子一抽想到这样一个式子：

\[\begin{align*}
&\sum_n f(n)\frac{1}{n!}x^n\\
=&\sum_{i=0}^m f_i\sum_{n\ge i} x^n\frac{1}{(n-i)!}\\
=&(\sum_{i=0}^m f_ix^i)e^x
\end{align*}
\]

于是\(f(x)\)在\([0,m]\)上的取值卷上一个\(e^{-x}\)就可以得到\(f_i\)，用NTT优化到\(m\log m\)。

脑洞大开，神仙题.jpg

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 100101
    #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,__zz=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[++__zz]=x%10+48,x/=10);
        while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
    }
    void file()
    {
        #ifdef NTFOrz
        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 r[sz],limit;
void NTT_init(int n)
{
	limit=1;int l=-1;
	while (limit<=n+n) limit<<=1,++l;
	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=a[j+k+mid]*w%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 n,X;
int m;

ll tmp1[sz],tmp2[sz],f[sz];

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

int main()
{
    file();
    init();
	read(n,m,X);
	rep(i,0,m) read(tmp1[i]),tmp1[i]=tmp1[i]*_fac[i]%mod;
	rep(i,0,m) tmp2[i]=((i&1)?mod-1:1ll)*_fac[i]%mod;
	NTT_init(m);
	NTT(tmp1,1);NTT(tmp2,1);
	rep(i,0,limit-1) f[i]=tmp1[i]*tmp2[i]%mod;
	NTT(f,-1);
	ll cur=1,ans=0;
	rep(i,0,m)
	{
		(ans+=f[i]*cur%mod)%=mod;
		cur=cur*(mod+n-i)%mod*X%mod;
	}
	cout<<ans;
	return 0;
}