正题

题目链接:https://www.luogu.com.cn/problem/P4717

---

题目大意

给出两个长度为$2^n$的数列$A,B$求
$$C_n=\sum _{iorj=n}{A}_i{B}_j$$
$$C_n=\sum _{iandj=n}{A}_i{B}_j$$
$$C_n=\sum _{ixorj=n}{A}_i{B}_j$$

---

解题思路

和$FFT$一样的思路，我们需要将$A,B$转换为点集表达式相乘再转回去，这里就直接抛结论了。

对于$or$有$$FWT(A)=(FWT(A),FWT(A)+FWT(B))$$

$$IFWT(A)=(IFWT(A),IFWT(A)-IFWT(B))$$
对于$and$有$$FWT(A)=(FWT(A)+FWT(B),FWT(B))$$

$$IFWT(A)=(IFWT(A)-IFWT(B),IFWT(B))$$
对于$xor$有$$FWT(A)=(FWT(A)+FWT(B),FWT(A)-FWT(B))$$

$$IFWT(A)=(\frac{IFWT(A)+IFWT(B)}{2},\frac{IFWT(A)-IFWT(B)}{2})$$
因为不用转成虚数所以代码好写很多，时间复杂度$O(n\log n)$

---

$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll inv2=499122177,N=(1<<18),P=998244353;
ll lg,n,aa[N],bb[N],a[N],b[N],c[3][N];void mul(ll *c,ll *a,ll *b){for(ll i=0;i<n;i++)c[i]=a[i]*b[i]%P;return;
}void FWT_or(ll *f,ll op){for(ll p=2;p<=n;p<<=1){ll len=p>>1;for(ll k=0;k<n;k+=p)for(ll i=k;i<k+len;i++)(f[i+len]+=f[i]*op+P)%=P;}return;
}
void solve_or(ll *c){memcpy(a,aa,sizeof(a));memcpy(b,bb,sizeof(b));FWT_or(a,1);FWT_or(b,1);mul(c,a,b); FWT_or(c,P-1);return;
}void FWT_and(ll *f,ll op){for(ll p=2;p<=n;p<<=1){ll len=p>>1;for(ll k=0;k<n;k+=p)for(ll i=k;i<k+len;i++)(f[i]+=f[i+len]*op+P)%=P;}return;
}
void solve_and(ll *c){memcpy(a,aa,sizeof(a));memcpy(b,bb,sizeof(b));FWT_and(a,1);FWT_and(b,1);mul(c,a,b); FWT_and(c,P-1);return;
}void FWT_xor(ll *f,ll op){for(ll p=2;p<=n;p<<=1){ll len=p>>1;for(ll k=0;k<n;k+=p)for(ll i=k;i<k+len;i++){ll x=f[i],y=f[i+len];f[i]=(x+y)*op%P;f[i+len]=(x-y+P)*op%P;}}return;
}
void solve_xor(ll *c){memcpy(a,aa,sizeof(a));memcpy(b,bb,sizeof(b));FWT_xor(a,1);FWT_xor(b,1);mul(c,a,b); FWT_xor(c,inv2);return;
}signed main()
{scanf("%lld",&lg);n=1<<lg;for(ll i=0;i<n;i++)scanf("%lld",&aa[i]);for(ll i=0;i<n;i++)scanf("%lld",&bb[i]);solve_or(c[0]);solve_and(c[1]);solve_xor(c[2]);for(ll i=0;i<n;i++)printf("%lld ",c[0][i]);putchar('\n');for(ll i=0;i<n;i++)printf("%lld ",c[1][i]);putchar('\n');for(ll i=0;i<n;i++)printf("%lld ",c[2][i]);putchar('\n');return 0;
}