UOJ269. 【清華集訓2016】如何優雅地求和 [生成函數]
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思路
神仙題.jpg
腦子一抽,想把\(f(x)\)表示成下降冪的形式,也就是
\[
f(x)=\sum_{i=0}^m f_ix_{(i)}\x_{(i)}=\prod_{k=0}^{i-1}(x-k)=[x\ge i]\frac{x!}{(x-i)!}
\]
這樣有什麽好處呢?回到原來的式子,我們有
\[
\begin{align*}
&\sum_{k=0}^n f(k){n\choose k}x^k(1-x)^{n-k}\=&\sum_{k=0}^n \sum_{i=0}^m f_ik_{(i)}{n\choose k}x^k(1-x)^{n-k}\=&\sum_{i=0}^mf_i \sum_{k=i}^n \frac{n!}{(n-k)!(k-i)!} x^k(1-x)^{n-k}\=&\sum_{i=0}^m f_i n_{(i)}\sum_{k=i}^n {n-i\choose k-i} x^{k}(1-x)^{n-k}\=&\sum_{i=0}^m f_in_{(i)}x^i
\end{align*}
\]
可以非常快地求出來。
還有一個問題:\(f_i\)怎麽求?
再一次腦子一抽想到這樣一個式子:
\[
\begin{align*}
&\sum_n f(n)\frac{1}{n!}x^n\=&\sum_{i=0}^m f_i\sum_{n\ge i} x^n\frac{1}{(n-i)!}\=&(\sum_{i=0}^m f_ix^i)e^x
\end{align*}
\]
於是\(f(x)\)在\([0,m]\)上的取值卷上一個\(e^x\)就可以得到\(f_i\),用NTT優化到\(m\log m\)。
腦洞大開,神仙題.jpg
代碼
#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; }
UOJ269. 【清華集訓2016】如何優雅地求和 [生成函數]