大意: 给定$N^3$立方体, 每个单位立方体权值为三个坐标异或, 每次沿坐标轴切一刀, 得分为两半内权值和的乘积, 求切成$n^3$块的最大得分.
可以发现得分与切法无关, 假设每个点权值为$a_i$, 就有$ans=\frac{(\sum a_i)^2-\sum a_i^2}{2}$.
从而转化为求$f_x=\sum\limits_{i=0}^{N-1}\sum\limits_{j=0}^{N-1}\sum\limits_{k=0}^{N-1}[(i\oplus j\oplus k)=x]$
可以用$FWT$很容易求出
#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
const int P = 998244353, INV2 = (P+1)/2;
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
const int N = 5e6+10;
int n, m, a[N]; void FWT(int *a, int n, int tp) {
for (int i=0; (1<<i)<n; ++i) {
REP(j,0,n-1) if (j>>i&1) {
int l = a[j^1<<i], r = a[j];
(a[j^1<<i] += r) %= P;
a[j] = (l-r)%P;
}
}
if (tp==-1) REP(i,0,n-1) a[i]=(ll)a[i]*inv(n)%P;
} int main() {
while (~scanf("%d", &n)) {
m = 1;
while (m<n) m *= 2;
REP(i,0,m-1) a[i]=i<n;
FWT(a,m,1);
REP(i,0,m-1) a[i]=(ll)a[i]*a[i]%P*a[i]%P;
FWT(a,m,-1);
int f1 = 0, f2 = 0;
REP(i,0,m-1) {
f1 = (f1+(ll)i*a[i])%P;
f2 = (f2+(ll)i*i%P*a[i])%P;
}
f1 = (ll)f1*f1%P;
int ans = (ll)(f1-f2)*INV2%P;
if (ans<0) ans += P;
printf("%d\n", ans);
}
}