NTT/FFT与字符串匹配

字符串匹配问题，除了可以用(KMP)、(AC)自动机等，有的还能利用(NTT/FFT)实现。
(Link)
本题中，对于每个在(B)中符合的位置(i)，都有(forall{jin[0,m-1]},a[j]=b[j+i]lor{a[j]='*'}lor{b[j+i]='*'})
注意到，(forall)一般用加法实现，(lor)一般用乘法实现。
具体的，在本题中，我们定义(F('*')=0,F(ch)=ch-'a'+1)，(xleftrightarrow{y})意为字符(x)可以与(y)匹配
那么(xleftrightarrow{y}=[F(x)*F(y)*(F(x)-F(y))^2=0])（一个为(0)，则等式成立）
则(forall{jin[0,m-1]},F(a_j)*F(b_{j+i})*(F(a_j)-F(b_{j+i}))^2=0)
即(sumlimits_{j=0}^{m-1}F(a_j)*F(b_{j+i})*(F(a_j)-F(b_{j+i}))^2=0)
令(a'_i=a_{n-i-1})，有(sumlimits_{j=0}^{m-1}F(a'_{n-j-1})*F(b_{j+i})*(F(a_{n-j-1})-F(b_{j+i}))^2=0)
转化成卷积形式，依次拆开，有：(res_{n+i-1}=sumlimits_{p+q=n+i-1}{a'_p}^3{b_q}+sumlimits_{p+q=n+i-1}{a'_p}{b_q}^3-2sumlimits_{p+q=n+i-1}{a'_p}^2{b_q}^2=0)
那么当(res_x=0)时，输出(x+2-n)（(+1)是因为题目下标从(1)开始）

Code

#include <bits/stdc++.h>

using namespace std;

const int mod = 998244353, g = 3, G = (mod + 1) / 3;

int n, m, t, k, tot, out[300005], p[1200005], a0[1200005], b0[1200005], a[1200005], b[1200005], res[1200005];

char s1[300005], s2[300005];

int read()
{
	int x = 0; char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
	return x;
}

int qpow(int base, int pw)
{
	int s = 1;
	while (pw)
	{
		if (pw & 1) s = 1ll * s * base % mod;
		base = 1ll * base * base % mod;
		pw >>= 1;
	}
	return s;
}

void g_l(int x)
{
	t = 1, k = 0;
	while (t <= x) t <<= 1, k ++ ;
	for (int i = 0; i < t; i ++ ) p[i] = (p[i >> 1] >> 1) | ((i & 1) << (k - 1));
	return;
}

void NTT(int *a, int o)
{
	for (int i = 0; i < t; i ++ ) if (i < p[i]) swap(a[i], a[p[i]]);
	for (int i = 1; i < t; i <<= 1)
	{
		int wn = qpow(o == 1 ? g : G, (mod - 1) / (i << 1));
		for (int j = 0; j < t; j += (i << 1))
		{
			int w = 1;
			for (int k = 0; k < i; k ++ , w = 1ll * w * wn % mod)
			{
				int p = a[j + k], q = 1ll * w * a[j + k + i] % mod;
				a[j + k] = (p + q) % mod, a[j + k + i] = (p - q + mod) % mod;
			}
		}
	}
	if (o == -1)
	{
		int div = qpow(t, mod - 2);
		for (int i = 0; i < t; i ++ ) a[i] = 1ll * a[i] * div % mod;
	}
	return;
}

int main()
{
	n = read(), m = read(), scanf("%s", s1), scanf("%s", s2);
	for (int i = 0; i < n; i ++ ) a0[i] = (s1[n - i - 1] == '*') ? 0 : s1[n - i - 1] - 'a' + 1;
	for (int i = 0; i < m; i ++ ) b0[i] = (s2[i] == '*') ? 0 : s2[i] - 'a' + 1;
	g_l(n + m);
	for (int i = 0; i < t; i ++ ) a[i] = 1ll * a0[i] * a0[i] % mod * a0[i] % mod, b[i] = b0[i];
	NTT(a, 1), NTT(b, 1);
	for (int i = 0; i < t; i ++ ) res[i] = (res[i] + 1ll * a[i] * b[i] % mod) % mod;
	for (int i = 0; i < t; i ++ ) a[i] = 1ll * a0[i] * a0[i] % mod, b[i] = 1ll * b0[i] * b0[i] % mod;
	NTT(a, 1), NTT(b, 1);
	for (int i = 0; i < t; i ++ ) res[i] = (res[i] - 2ll * a[i] * b[i] % mod + mod * 2ll) % mod;
	for (int i = 0; i < t; i ++ ) a[i] = a0[i], b[i] = 1ll * b0[i] * b0[i] % mod * b0[i] % mod;
	NTT(a, 1), NTT(b, 1);
	for (int i = 0; i < t; i ++ ) res[i] = (res[i] + 1ll * a[i] * b[i] % mod) % mod;
	NTT(res, -1);
	for (int i = n - 1; i <= m - 1; i ++ ) if (!res[i]) out[ ++ tot] = i + 2 - n;
	printf("%d
", tot); for (int i = 1; i <= tot; i ++ ) printf("%d ", out[i]); putchar('
');
	return 0;
}