CF923E Perpetual Subtraction
考虑每次转移前后的关系：
令 \(\displaystyle F(x)=\sum_{i=0}^nf_{i}x^i\) , \(F^*(x)\) 为操作后的生成函数。
\[\begin{aligned}
F^*(x)&= \sum_{i=0}^n x^i\sum_{j=i}^n \frac{f_j}{j+1}\\
&= \sum_{i=0}^n \frac{f_i}{i+1}\sum_{j=0}^i x^j\\
&=\sum_{i=0}^n\frac{f_i}{i+1}\frac{1-x^{i+1}}{1-x} \\
&=\frac{1}{1-x}\sum_{i=0}^n\frac{f_i}{i+1}\left(1-x^{i+1} \right) \\
&=\frac{1}{x-1}\sum_{i=0}^n f_i\left(\frac{x^{i+1}-1^{i+1}}{i+1} \right) \\
&=\frac{1}{x-1}\sum_{i=0}^n f_i \int_{1}^x t^i ~\text{d}t \\
&=\frac{\int_{1}^x F(t)~ \text{d}t}{x-1}
\end{aligned}\]
有一个自然的想法: 令 \(G(x-1)=F(x)\)，那么有：
\[G^*(x)=F^*(x+1)=\frac{\int_0^xG(t) ~\text{d} t}{x}
\]
又因为：
\[\int G(x)=\sum_{i=0}^n\frac{g_i}{i+1}x^{i+1}+C
\]
得到：
\[G^*(x)=\sum_{i=0}^n \frac{g_i}{i+1}x^i
\]
那么：
\[G_m(x)=\sum_{i=0}^n \frac{g_{0,i}}{(i+1)^m}x^i
\]
显然有 \(\displaystyle F_0(x)=\sum_{i=0}^n p_ix^i\)
只需要知道 \(F,G\) 之间如何转换即可。
- \(F \to G\)
\[\begin{aligned}
G(x)&=F(x+1) \\
&=\sum_{i=0}^n f_i(x+1)^i \\
&=\sum_{i=0}^n f_i\sum_{j=0}^i \binom{i}{j} x^j\\
&=\sum_{j=0}^n x^j \sum_{i=j}^n \binom{i}{j}f_i \\
&=\sum_{j=0}^n x^j\frac{1}{j!}\sum_{i=j}^n \frac{f_ii!}{(i-j)!}
\end{aligned}\]
反转 \(f_ii!\) 后便可得到卷积形式。
- \(G \to F\)
\[\begin{aligned}
F(x)&=G(x-1) \\
&=\sum_{i=0}^n g_i(x-1)^i \\
&=\sum_{i=0}^n g_i\sum_{j=0}^i \binom{i}{j} x^j(-1)^{i-j}\\
&=\sum_{j=0}^n x^j \sum_{i=j}^n \binom{i}{j}g_i (-1)^{i-j}\\
&=\sum_{j=0}^n x^j\frac{1}{j!}\sum_{i=j}^n g_ii!\frac{(-1)^{i-j}}{(i-j)!}
\end{aligned}\]
同样反转 \(g_i i!\) 为卷积形式。
那么 \(F_0 \Rightarrow G_0\) , \(G_0 \Rightarrow G_m , G_m \Rightarrow F_m\) 即可
两次卷积转换 \(f,g\) ，时间复杂度 \(\mathcal{ O(n \log n ) }\)
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
const int MAXN = 4e5 , Mod = 998244353;
int Add( int x , int y ) { x += y; return x >= Mod ? x - Mod : x; }
int Sub( int x , int y ) { x -= y; return x < 0 ? x + Mod : x; }
int Mul( int x , int y ) { return 1ll * x * y % Mod; }
int Quick_pow( int x , int po ) { int Ans = 1; for( ; po ; po >>= 1 , x = Mul( x , x ) ) if( po & 1 ) Ans = Mul( Ans , x ); return Ans; }
int Inv( int x ) { return Quick_pow( x , Mod - 2 ); }
int fac[ MAXN + 5 ] , ivf[ MAXN + 5 ];
void Init( ) {
fac[ 0 ] = 1;
for( int i = 1 ; i <= MAXN ; i ++ ) fac[ i ] = Mul( fac[ i - 1 ] , i );
ivf[ MAXN ] = Inv( fac[ MAXN ] );
for( int i = MAXN ; i >= 1 ; i -- ) ivf[ i - 1 ] = Mul( ivf[ i ] , i );
}
#define Poly vector< int >
#define len( x ) ( (int)x.size() )
const int G = 3 , IG = 332748118;
int lim , ilim , rev[ MAXN + 5 ];
void ntt( Poly &f , int op ) {
for( int i = 0 ; i < lim ; i ++ ) if( i < rev[ i ] ) swap( f[ i ] , f[ rev[ i ] ] );
for( int len = 2 , w ; len <= lim ; len <<= 1 ) {
w = Quick_pow( op == 1 ? G : IG , ( Mod - 1 ) / len );
for( int l = 0 ; l < lim ; l += len ) {
for( int i = l , wk = 1 ; i < l + len / 2 ; i ++ , wk = Mul( wk , w ) ) {
int t = Mul( wk , f[ i + len / 2 ] );
f[ i + len / 2 ] = Sub( f[ i ] , t ); f[ i ] = Add( f[ i ] , t );
}
}
}
if( op == -1 ) for( int i = 0 ; i < lim ; i ++ ) f[ i ] = Mul( f[ i ] , ilim );
}
Poly operator * ( Poly f , Poly g ) {
int n = len( f ) + len( g ) - 1; for( lim = 1 ; lim < n ; lim <<= 1 ); ilim = Inv( lim );
for( int i = 0 ; i < lim ; i ++ ) rev[ i ] = ( rev[ i >> 1 ] >> 1 ) | ( i & 1 ? lim >> 1 : 0 );
f.resize( lim ); g.resize( lim );
ntt( f , 1 ); ntt( g , 1 );
for( int i = 0 ; i < lim ; i ++ ) f[ i ] = Mul( f[ i ] , g[ i ] );
ntt( f , -1 ); f.resize( n );
return f;
}
int n; LL m;
Poly Solve( Poly f , int op ) {
Poly g; g.resize( n + 1 );
for( int i = 0 ; i <= n ; i ++ ) f[ i ] = Mul( f[ i ] , fac[ i ] );
reverse( f.begin() , f.end() );
for( int i = 0 ; i <= n ; i ++ ) g[ i ] = ( ( i & 1 ) && ( op == -1 ) ) ? Mod - ivf[ i ] : ivf[ i ];
f = f * g; f.resize( n + 1 );
reverse( f.begin() , f.end() );
for( int i = 0 ; i <= n ; i ++ ) f[ i ] = Mul( f[ i ] , ivf[ i ] );
return f;
}
Poly f;
int main( ) {
Init();
scanf("%d %lld",&n,&m); f.resize( n + 1 );
for( int i = 0 ; i <= n ; i ++ ) scanf("%d", &f[ i ] );
f = Solve( f , 1 );
for( int i = 0 ; i <= n ; i ++ ) f[ i ] = Mul( f[ i ] , Inv( Quick_pow( i + 1 , m % ( Mod - 1 ) ) ) );
f = Solve( f , -1 );
for( int i = 0 ; i <= n ; i ++ ) printf("%d%c", f[ i ] , i == n + 1 ? '\n' : ' ' );
return 0;
}
