【Codeforces 1119H】Triple（FWT）

传送门

和黎明前的巧克力没什么区别
只是有三个变量了
考虑先对于$a,b,c$变成$0,boplus a,coplus a$
最后异或$oplus_i a_i$就可以了
这样就只有2个了
做三次$fwt$即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
cs int mod=998244353;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b%mod;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b%mod,a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline void fix(int &x){(x>mod-1)&&(x-=mod-1),(x<0)&&(x+=mod-1);}
cs int N=(1<<18)|1,iv4=Inv(4);
int f1[N],f2[N],f3[N];
int n,k,sta;
inline void dft(int *f,int lim,int kd){
	for(int mid=1,a0,a1;mid<lim;mid<<=1)
	for(int i=0;i<lim;i+=mid<<1)
	for(int j=0;j<mid;j++)
	a0=f[i+j],a1=f[i+j+mid],f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	if(kd==-1){
		for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
	}
}
int x,y,z;
int v[N];
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	n=read(),k=read(),sta=1<<k;
	int allxor=0;
	x=read()%mod,y=read()%mod,z=read()%mod;
	for(int i=1;i<=n;i++){
		int a=read(),b=read()^a,c=read()^a;
		allxor^=a;
		f1[b]++,f2[c]++,f3[b^c]++;
	}
	dft(f1,sta,1),dft(f2,sta,1),dft(f3,sta,1);
	int k1=add(x,add(y,z)),k2=add(x,dec(y,z)),k3=add(dec(x,y),z),k4=dec(dec(x,y),z);
	for(int i=0;i<sta;i++){
		int v1=f1[i],v2=f2[i],v3=f3[i];
		int x1=mul(iv4,add(add(n,v1),add(v2,v3))),x2=mul(iv4,dec(add(n,v1),add(v2,v3))),
		x3=mul(iv4,dec(add(n,v2),add(v1,v3))),x4=mul(iv4,dec(add(n,v3),add(v1,v2)));
		v[i]=mul(mul(ksm(k1,x1),ksm(k2,x2)),mul(ksm(k3,x3),ksm(k4,x4)));
	}
	dft(v,sta,-1);
	for(int i=0;i<sta;i++)
	cout<<v[i^allxor]<<" ";
	return 0;
}