FWT 快速沃尔什变换

目前我了解的FWT的功能就是在 $O(nlog n) 时间复杂度内求：$

$C_k = sum_{i|j=k}a_ib_j$

$C_k = sum_{i&j=k}a_ib_j$

$C_k = sum_{ioplus j=k}a_ib_j$

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其思想和 $FFT$ 很类似。把原多项式 $A$ 转化成另一个多项式 $FWT(A)$。

然后 $A oplus B = IFWT(FWT(A) times FWT(B))$

其中 $oplus$ 表示一些位运算。

然后接下来的问题是针对不同的位运算，去找出如何求 $FWT$ 和 $IFWT$。

一些定义

$A = [a_0, a_1, a_2, …, a_{n-1}]$

设 $A$ 的长度为 $2$ 的幂次，将 $A = [A_0, A_1]$ 分成长度相同的两部分。

OR

$C = A|B = [sum_{i|j=0}a_ib_j, sum_{i|j=1}a_ib_j, sum_{i|j=2}a_ib_j, …, sum_{i|j=k}a_ib_j]$

$C_k = sum_{i|j=k}a_ib_j$

$FWT(A) = [sum_{i | 0 = 0}a_i, sum_{i | 1 = 1}a_i, sum_{i | 2 = 2}a_i, …, sum_{i | (n-1) = (n-1)}a_{n-1}]$

FWT 变换所满足的规律

$$FWT(A) = [FWT(A_0), FWT(A_0 + A_1)]$$

FWT 变换所满足的性质

$FWT(A+B) = FWT(A) + FWT(B)$

$FWT(A|B) = FWT(A) times FWT(B)$

$FWT(A|B) = FWT([(A|B)_0, (A|B)_1])$
$ = [FWT((A|B)_0), FWT((A|B)_0 + (A|B)_1)]$
$ = [FWT(A_0 | B_0), FWT(A_0 | B_0 + A_0 | B_1 + A_1 | B_0 + A_1 | B_1)]$
$ = [FWT(A_0) times FWT(B_0), FWT(A_0) times FWT(B_0) + FWT(A_0) times FWT(B_1) + FWT(A_1) times FWT(B_0) + FWT(A_1) times FWT(B_1))]$
$ = [FWT(A_0) times FWT(B_0), (FWT(A_0) + FWT(A_1)) times (FWT(B_0) + FWT(B_1))]$
$ = [FWT(A_0), FWT(A_0 + A_1)] times [FWT(B_0), FWT(B_0 + B_1)]$
$ = FWT(A) times FWT(B)$

IFWT 变换所满足的规律

$$IFWT(A) = [IFWT(A_0), IFWT(A_1) - IFWT(A_0)]$$

AND

$C = A&B = [sum_{i&j=0}a_ib_j, sum_{i&j=1}a_ib_j, sum_{i&j=2}a_ib_j, …, sum_{i&j=k}a_ib_j]$

$C_k = sum_{i&j=k}a_ib_j$

FWT 变换所满足的规律

$$FWT(A) = [FWT(A_0 + A_1), FWT(A_1)]$$

FWT 变换所满足的性质

$FWT(A+B) = FWT(A) + FWT(B)$

$FWT(A&B) = FWT(A) times FWT(B)$

IFWT 变换所满足的规律

$$IFWT(A) = [IFWT(A_0) - IFWT(A_1), IFWT(A_1)]$$

XOR

$C = Aoplus B = [sum_{ioplus j=0}a_ib_j, sum_{ioplus j=1}a_ib_j, sum_{ioplus j=2}a_ib_j, …, sum_{ioplus j=k}a_ib_j]$

$C_k = sum_{ioplus j=k}a_ib_j$

FWT 变换所满足的规律

$$FWT(A) = [FWT(A_0) + FWT(A_1), FWT(A_0) - FWT(A_1)]$$

FWT 变换所满足的性质

$FWT(A+B) = FWT(A) + FWT(B)$

$FWT(Aoplus B) = FWT(A) times FWT(B)$

IFWT 变换所满足的规律

$$IFWT(A) = [frac{IFWT(A_0) + IFWT(A_1)}{2}, frac{IFWT(A_0) - IFWT(A_1)}{2}$$

代码

洛谷P4717

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using namespace std; const int mod = 998244353; const int maxn = 1e6 + 6; int a[maxn], b[maxn];大专栏  FWT 快速沃尔什变换an> int inv2 = (mod + 1) / 2; template <class > void fwt(int a[], int m, f) { for (int i = 1; i < m; i <<= 1) for (int p = i << 1, j = 0; j < m; j += p) for (int k = 0; k < i; k++) f(a[j + k], a[i + j + k]); } inline void AND(int &a, int &b) { (a += b) %= mod; } inline void OR(int &a, int &b) { (b += a) %= mod; } inline void XOR(int &a, int &b) { int l = a, r = b; a = (l + r) % mod, b = (l - r + mod) % mod; } inline void rAND(int &a, int &b) { a = (a - b + mod) % mod; } inline void rOR(int &a, int &b) { b = (b - a + mod) % mod; } inline void rXOR(int &a, int &b) { int l = a, r = b; a = 1LL * (l + r) % mod * inv2 % mod; b = 1LL * (l - r + mod) % mod * inv2 % mod; } int c[maxn], d[maxn]; int n, N; void solve(int a[], int b[], int op) { for (int i = 0; i < N; i++) c[i] = a[i], d[i] = b[i]; if (op == 0) { fwt(c, N, OR); fwt(d, N, OR); for (int i = 0; i < N; i++) c[i] = 1LL * c[i] * d[i] % mod; fwt(c, N, rOR); } else if (op == 1) { fwt(c, N, AND); fwt(d, N, AND); for (int i = 0; i < N; i++) c[i] = 1LL * c[i] * d[i] % mod; fwt(c, N, rAND); } else { fwt(c, N, XOR); fwt(d, N, XOR); for (int i = 0; i < N; i++) c[i] = 1LL * c[i] * d[i] % mod; fwt(c, N, rXOR); } for (int i = 0; i < N; i++) printf("%d ", c[i]); puts(""); } int main() { scanf("%d", &n); N = 1 << n; for (int i = 0; i < N; i++) scanf("%d", a + i); for (int i = 0; i < N; i++) scanf("%d", b + i); solve(a, b, 0); solve(a, b, 1); solve(a, b, 2); }