[洛谷P4717]【模板】快速沃尔什变换

题目大意：给定多项式$A$和$B$，求$C$满足：

$$
C_n=\sum\limits_{x\oplus y=n}A_xB_y
$$

其中$\oplus$为位运算（$or,and,xor​$）

题解：$FWT$，可以见这篇博客

卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#define maxn 262144
const int mod = 998244353;

namespace Math {
	inline int pw(int base, int p) {
		static int res;
		for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	inline int inv(int x) { return pw(x, mod - 2); }
}

inline void reduce(int &x) { x += x >> 31 & mod; }
inline void clear(register int *l, const int *r) {
	if (l >= r) return ;
	while (l != r) *l++ = 0;
}

int __n, n, lim;
int A[maxn], B[maxn], C[maxn], D[maxn];

inline void FWT_OR(int *A) {
	for (register int mid = 1; mid < lim; mid <<= 1)
		for (register int i = 0; i < lim; i += mid << 1)
			for (register int j = 0; j < mid; ++j) reduce(A[i + j + mid] += A[i + j] - mod);
}
inline void IFWT_OR(int *A) {
	for (register int mid = 1; mid < lim; mid <<= 1)
		for (register int i = 0; i < lim; i += mid << 1)
			for (register int j = 0; j < mid; ++j) reduce(A[i + j + mid] -= A[i + j]);
}

inline void FWT_AND(int *A) {
	for (register int mid = 1; mid < lim; mid <<= 1)
		for (register int i = 0; i < lim; i += mid << 1)
			for (register int j = 0; j < mid; ++j) reduce(A[i + j] += A[i + j + mid] - mod);
}
inline void IFWT_AND(int *A) {
	for (register int mid = 1; mid < lim; mid <<= 1)
		for (register int i = 0; i < lim; i += mid << 1)
			for (register int j = 0; j < mid; ++j) reduce(A[i + j] -= A[i + j + mid]);
}

inline void FWT_XOR(int *A, const int op = 1) {
	for (register int mid = 1; mid < lim; mid <<= 1)
		for (register int i = 0; i < lim; i += mid << 1)
			for (register int j = 0; j < mid; ++j) {
				const int X = A[i + j], Y = A[i + j + mid];
				reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
			}
	if (!op) {
		const int ilim = Math::inv(lim);
		for (register int *i = A; i != A + lim; ++i) *i = static_cast<long long> (*i) * ilim % mod;
	}
}

int main() {
	scanf("%d", &__n); n = Math::pw(2, __n);
	for (int i = 0; i < n; ++i) scanf("%d", A + i);
	for (int i = 0; i < n; ++i) scanf("%d", B + i);
	lim = n;

	std::copy(A, A + n, C); clear(C + n, C + lim);
	std::copy(B, B + n, D); clear(D + n, D + lim);
	FWT_OR(C), FWT_OR(D);
	for (int i = 0; i < lim; ++i) C[i] = static_cast<long long> (C[i]) * D[i] % mod;
	IFWT_OR(C);
	for (int i = 0; i < n; ++i) printf("%d ", C[i]); puts("");

	std::copy(A, A + n, C); clear(C + n, C + lim);
	std::copy(B, B + n, D); clear(D + n, D + lim);
	FWT_AND(C), FWT_AND(D);
	for (int i = 0; i < lim; ++i) C[i] = static_cast<long long> (C[i]) * D[i] % mod;
	IFWT_AND(C);
	for (int i = 0; i < n; ++i) printf("%d ", C[i]); puts("");

	std::copy(A, A + n, C); clear(C + n, C + lim);
	std::copy(B, B + n, D); clear(D + n, D + lim);
	FWT_XOR(C), FWT_XOR(D);
	for (int i = 0; i < lim; ++i) C[i] = static_cast<long long> (C[i]) * D[i] % mod;
	FWT_XOR(C, 0);
	for (int i = 0; i < n; ++i) printf("%d ", C[i]); puts("");
	return 0;
}