题面

Sol

假设没有通配符

那么把\(T\)翻转

设\(f[i]=\sum_{j+k=i}[S[k]==T[j]]\)

如果\(f[i]\)为\(0\)则\(i\)之前的一一匹配

那么可以给每个字符一个权值

重新定义\(f[i]=\sum_{j+k=i}(S[k]-T[j])^2\)

就可以\(FFT\)了

然后有通配符，设权值为\(0\)

再定义

\(f[i]=\sum_{j+k=i}(S[k]-T[j])^2S[k]T[j]\)

然后拆开\(FFT\)就可以了

# include 
    
     # define IL inline# define RG register# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;template 
     
      IL void Input(RG Int &x){    RG int z = 1; RG char c = getchar(); x = 0;    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    x *= z;}const int maxn(3e5 + 5);const int oo(1e9);const double pi(acos(-1));int n, m, k, r[maxn << 2], len, cnt, v1[maxn], v2[maxn], ans[maxn];char s[maxn], t[maxn];struct Complex{    double real, image;    IL Complex(){        real = image = 0;    }        IL Complex(RG double a, RG double b){        real = a, image = b;    }    IL Complex operator +(RG Complex b){        return Complex(real + b.real, image + b.image);    }    IL Complex operator -(RG Complex b){        return Complex(real - b.real, image - b.image);    }    IL Complex operator *(RG Complex b){        return Complex(real * b.real - image * b.image, real * b.image + image * b.real);    }    IL Complex operator *(RG int b){        return Complex(real * b, image * b);    }} a[maxn << 2], b[maxn << 2], w[maxn << 2], c[maxn << 2];IL void FFT(RG Complex *p, RG int opt){    for(RG int i = 0; i < len; ++i) if(r[i] < i) swap(p[i], p[r[i]]);    for(RG int i = 1; i < len; i <<= 1)        for(RG int j = 0, l = i << 1; j < len; j += l){            for(RG int k = 0; k < i; ++k){                RG Complex wn = Complex(w[len / i * k].real, w[len / i * k].image * opt);                RG Complex x = p[k + j], y = wn * p[k + j + i];                p[k + j] = x + y, p[k + j + i] = x - y;            }        }}IL void Prepare(){    RG int l = 0, tmp = m + n - 1;    for(len = 1; len < tmp; len <<= 1) ++l;    for(RG int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));    for(RG int i = 1; i <= len; i <<= 1)        for(RG int k = 0; k < i; ++k)            w[len / i * k] = Complex(cos(pi / i * k), sin(pi / i * k));}int main(RG int argc, RG char* argv[]){    Input(m), Input(n), scanf(" %s %s", t, s);    reverse(t, t + m), Prepare();    for(RG int i = 0; i < n; ++i) v1[i] = s[i] == '*' ? 0 : s[i] - 'a' + 1;    for(RG int i = 0; i < m; ++i) v2[i] = t[i] == '*' ? 0 : t[i] - 'a' + 1;        for(RG int i = 0; i < n; ++i) a[i].real = v1[i] * v1[i] * v1[i];    for(RG int i = 0; i < m; ++i) b[i].real = v2[i];    FFT(a, 1), FFT(b, 1);    for(RG int i = 0; i < len; ++i) c[i] = c[i] + a[i] * b[i], a[i] = b[i] = Complex(0, 0);        for(RG int i = 0; i < n; ++i) a[i].real = v1[i] * v1[i] << 1;    for(RG int i = 0; i < m; ++i) b[i].real = v2[i] * v2[i];    FFT(a, 1), FFT(b, 1);    for(RG int i = 0; i < len; ++i) c[i] = c[i] - a[i] * b[i], a[i] = b[i] = Complex(0, 0);        for(RG int i = 0; i < n; ++i) a[i].real = v1[i];    for(RG int i = 0; i < m; ++i) b[i].real = v2[i] * v2[i] * v2[i];    FFT(a, 1), FFT(b, 1);    for(RG int i = 0; i < len; ++i) c[i] = c[i] + a[i] * b[i];        FFT(c, -1);    for(RG int i = m - 1; i < n; ++i)        if(!(int(c[i].real / len + 0.5))) ans[++cnt] = i;    printf("%d\n", cnt);    for(RG int i = 1; i <= cnt; ++i) printf("%d ", ans[i] - m + 2);    return puts(""), 0;}