前言
学多项式怎么能错过\(FWT\)呢,然而这真是个毒瘤的东西,蒟蒻就只会背公式了\(\%>\_<\%\)
或卷积
\[\begin{aligned}\\
tf(A) = (tf(A_0), tf(A_1) + tf(A_0))\\
utf(A) = (utf(A), utf(A_1) - utf(A_0))\\
\end{aligned}\]
tf(A) = (tf(A_0), tf(A_1) + tf(A_0))\\
utf(A) = (utf(A), utf(A_1) - utf(A_0))\\
\end{aligned}\]
与卷积
\[\begin{aligned}\\
tf(A) = (tf(A_0) + tf(A_1), tf(A_1))\\
utf(A) = (utf(A_0) - utf(A_1), utf(A_1))\\
\end{aligned}\]
tf(A) = (tf(A_0) + tf(A_1), tf(A_1))\\
utf(A) = (utf(A_0) - utf(A_1), utf(A_1))\\
\end{aligned}\]
异或卷积
\[\begin{aligned}\\
tf(A) = (tf(A_0) + tf(A_1), tf(A_0) - tf(A_1))\\
utf(A) = (\frac{utf(A_0) + utf(A_1)}{2}, \frac{utf(A_0) - utf(A_1)}{2})\\
\end{aligned}\]
tf(A) = (tf(A_0) + tf(A_1), tf(A_0) - tf(A_1))\\
utf(A) = (\frac{utf(A_0) + utf(A_1)}{2}, \frac{utf(A_0) - utf(A_1)}{2})\\
\end{aligned}\]
Code
习惯写递归的非递归本来也不会
#include<bits/stdc++.h>
typedef int LL;
inline LL Read(){
LL x(0),f(1); 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-'0'; c=getchar();
}
return x*f;
}
const LL mod=998244353,maxn=1<<18,inv2=499122177;
inline LL Pow(LL base,LL b){
LL ret(1);
while(b){
if(b&1) ret=1ll*ret*base%mod; base=1ll*base*base%mod; b>>=1;
}
return ret;
}
void Solve_or(LL n,LL *a,LL *b,LL *c){
n>>=1;
if(!n){
c[0]=1ll*a[0]*b[0]%mod;
return;
}
for(LL i=0;i<n;++i){
a[i+n]=1ll*(a[i+n]+a[i])%mod; b[i+n]=1ll*(b[i+n]+b[i])%mod;
}
Solve_or(n,a,b,c); Solve_or(n,a+n,b+n,c+n);
for(LL i=0;i<n;++i) c[i+n]=(c[i+n]-c[i]+mod)%mod;
}
void Solve_and(LL n,LL *a,LL *b,LL *c){
n>>=1;
if(!n){
c[0]=1ll*a[0]*b[0]%mod;
return;
}
for(LL i=0;i<n;++i){
a[i]=1ll*(a[i]+a[i+n])%mod; b[i]=1ll*(b[i]+b[i+n])%mod;
}
Solve_and(n,a,b,c); Solve_and(n,a+n,b+n,c+n);
for(LL i=0;i<n;++i) c[i]=1ll*(c[i]-c[i+n]+mod)%mod;
}
void Solve_xor(LL n,LL *a,LL *b,LL *c){
n>>=1;
if(!n){
c[0]=1ll*a[0]*b[0]%mod;
return;
}
for(LL i=0;i<n;++i){
std::tie(a[i],a[i+n])=std::make_tuple(a[i]+a[i+n],a[i]-a[i+n]+mod);
std::tie(b[i],b[i+n])=std::make_tuple(b[i]+b[i+n],b[i]-b[i+n]+mod);
a[i]%=mod; a[i+n]%=mod; b[i]%=mod; b[i+n]%=mod;
}
Solve_xor(n,a,b,c); Solve_xor(n,a+n,b+n,c+n);
for(LL i=0;i<n;++i){
std::tie(c[i],c[i+n])=std::make_tuple(c[i]+c[i+n],c[i]-c[i+n]+mod);
c[i]=1ll*c[i]%mod*inv2%mod; c[i+n]=1ll*c[i+n]%mod*inv2%mod;
}
}
LL n,N;
LL a[maxn],b[maxn],c[maxn],d[maxn],e[maxn],f[maxn],x[maxn],y[maxn],z[maxn];
int main(){
n=Read();
N=1<<n;
for(LL i=0;i<N;++i) a[i]=c[i]=e[i]=Read();
for(LL i=0;i<N;++i) b[i]=d[i]=f[i]=Read();
Solve_or(N,a,b,x);
Solve_and(N,c,d,y);
Solve_xor(N,e,f,z);
for(LL i=0;i<N;++i) printf("%d ",x[i]);printf("\n");
for(LL i=0;i<N;++i) printf("%d ",y[i]);printf("\n");
for(LL i=0;i<N;++i) printf("%d ",z[i]);printf("\n");
return 0;
}