链式反应

题目描述

不想看题面，其实就是给定$p(x)$，有$f^{\prime }(x)=f(x{)}^{2}p(x)+1$，求$f$这个多项式的幂级数形式前$n$项。

Solution

式子可以写成$f_n={\sum }_{i,j}[0<=i+j<n]f_i{f}_j{p}_{n-i-j-1}$。

显然能分治$FFT$。
设分治区间为$[l,r]$，考虑$[l,mid]$对$[mid+1,r]$的贡献。
当$l=1$时，贡献为
$$\sum _{i=0}^{mid}\sum _{j=0}^{mid}\sum _{k=0}^r{f}_i{f}_j{p}_k$$
当$l>1$时，有$2l>r$，因此贡献为
$$2\sum _{i=l}^{mid}\sum _{j=0}^{r-l}\sum _{k=0}^{r-l}{f}_i{f}_j{p}_k$$
时间复杂度两只$lg$。
注意这里的$f,p$始终是一个$EGF$。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int g=3;
const int gi=(mods+1)/3;
const int MAXN=2000005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int Limit,L;
char st[MAXN];
int F[MAXN],G[MAXN],P[MAXN],H[MAXN],Q[MAXN],rev[MAXN],inv[MAXN],fac[MAXN],n;
inline int upd(int x,int y) { return (x+y>=mods)?x+y-mods:x+y; }
inline int quick_pow(int x,int y)
{int ret=1;for (;y;y>>=1){if (y&1) ret=1ll*ret*x%mods;x=1ll*x*x%mods;}return ret;
}
inline void Init(int n)
{fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;inv[n]=quick_pow(fac[n],mods-2);for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mods;
}
void Number_Theoretic_Transform(int *A,int opt)
{for (int i=0;i<Limit;i++) if (i<rev[i]) swap(A[i],A[rev[i]]);for (int i=1;i<Limit;i<<=1){int Wn=quick_pow(opt==1?g:gi,(mods-1)/(i<<1));for (int j=0;j<Limit;j+=(i<<1))for (int k=j,w=1;k<j+i;k++,w=1ll*w*Wn%mods){int x=A[k],y=1ll*A[k+i]*w%mods;A[k]=upd(x,y),A[k+i]=upd(x,mods-y);}}if (opt==-1){int invlim=quick_pow(Limit,mods-2);for (int i=0;i<Limit;i++) A[i]=1ll*A[i]*invlim%mods;}
}
void solve(int l,int r)
{if (l==r) { if (l==1) F[l]=(mods+1)>>1;else F[l]=1ll*F[l]*inv[l]%mods*fac[l-1]%mods;printf("%d\n",2ll*F[l]*fac[l]%mods);return; }int mid=(l+r)>>1;solve(l,mid);if (l==1){Limit=1,L=0;int len=mid*2+r;while (Limit<=len) Limit<<=1,L++;for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));for (int i=0;i<Limit;i++) H[i]=G[i]=0;for (int i=0;i<=r;i++) H[i]=P[i];for (int i=0;i<=mid;i++) G[i]=F[i];Number_Theoretic_Transform(G,1);Number_Theoretic_Transform(H,1);for (int i=0;i<Limit;i++) G[i]=1ll*G[i]*G[i]%mods*H[i]%mods;Number_Theoretic_Transform(G,-1);for (int i=mid+1;i<=r;i++) F[i]=upd(F[i],G[i]);}else{Limit=1,L=0;int len=(mid-l)+(r-l)*2;while (Limit<=len) Limit<<=1,L++;for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));for (int i=0;i<Limit;i++) H[i]=G[i]=Q[i]=0;for (int i=0;i<=r-l;i++) H[i]=P[i];for (int i=0;i<=mid-l;i++) G[i]=F[i+l];for (int i=0;i<=r-l;i++) Q[i]=F[i];Number_Theoretic_Transform(G,1);Number_Theoretic_Transform(Q,1);Number_Theoretic_Transform(H,1);for (int i=0;i<Limit;i++) G[i]=1ll*G[i]*Q[i]%mods*H[i]%mods;Number_Theoretic_Transform(G,-1);for (int i=mid+1;i<=r;i++) F[i]=upd(F[i],upd(G[i-l],G[i-l]));}solve(mid+1,r);
}
int main()
{n=read();Init(n);scanf("%s",st);for (int i=1;i<=n;i++) P[i]=(st[i-1]-'0')*inv[i-1];solve(1,n);return 0;
}