2019杭电多校二 F. Fantastic Magic Cube (FWT)

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$f_x=sumlimits_{i=0}^{N-1}sumlimits_{j=0}^{N-1}sumlimits_{k=0}^{N-1}[(ioplus joplus k)=x]$

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
", ans);
	}
}