bzoj4259

fft

搞一个生成函数

对于每位A(j)=Σi=1->m (a[i]-b[i+j])^2*a[i]*b[i+j]

如果A(j)=0说明这位匹配

如果这位是*那么a[i]=0否则等于字母-'a'+1,b也是这样构造

然后我们翻转a串就可以加速了

#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1)
const int N = ( << ) + ;
int n = , n1, n2, k;
char s1[N], s2[N];
int t[N], ans[N];
struct data {
    double a, b;
    data() { a = ; b = ; }
    data(double _, double __) : a(_), b(__) {}
    data friend operator + (const data &a, const data &b) { return data(a.a + b.a, a.b + b.b); }
    data friend operator - (const data &a, const data &b) { return data(a.a - b.a, a.b - b.b); }
    data friend operator * (const data &a, const data &b) { return data(a.a * b.a - a.b * b.b, a.a * b.b + a.b * b.a); }
} a0[N], b0[N], a1[N], b1[N], a2[N], b2[N];
void fft(data *a, int f)
{
    for(int i = ; i < n; ++i)
    {
        int t = ;
        for(int j = ; j < k; ++j) if(i >> j & ) t |=  << (k - j - );
        if(i < t) swap(a[i], a[t]);
    }
    for(int l = ; l <= n; l <<= )
    {
        int m = l >> ;
        data w = data(cos(pi / m), f * sin(pi / m));
        for(int i = ; i < n; i += l)
        {
            data t = data(, );
            for(int k = ; k < m; ++k, t = t * w)
            {
                data x = a[i + k], y = t * a[i + k + m];
                a[i + k] = x + y;
                a[i + k + m] = x - y;
            }
        }
    }
}
int main()
{
    scanf("%d%d%s%s", &n1, &n2, s1, s2);
    reverse(s1, s1 + n1);
    --n1;
    --n2;
    for(int i = ; i <= n1; ++i) if(s1[i] != '*')
    {
        double x = s1[i] - 'a' + ;
        a0[i] = data(x, );
        a1[i] = data(x * x, );
        a2[i] = data(x * x * x, );
    }
    for(int i = ; i <= n2; ++i) if(s2[i] != '*')
    {
        double x = s2[i] - 'a' + ;
        b0[i] = data(x, );
        b1[i] = data(- * x * x, );
        b2[i] = data(x * x * x, );
    }
    while(n <= n1 + n2) n <<= , ++k;
    fft(a0, );
    fft(a1, );
    fft(a2, );
    fft(b0, );
    fft(b1, );
    fft(b2, );
    for(int i = ; i < n; ++i) a2[i] = a2[i] * b0[i], a1[i] = a1[i] * b1[i], a0[i] = a0[i] * b2[i], a2[i] = a2[i] + a1[i] + a0[i];
    fft(a2, -);
    for(int i = ; i < n; ++i) t[i] = (int)(a2[i].a / n + 0.1);
    for(int i = ; i <= n2 - n1; ++i) if(t[i + n1] == ) ans[++ans[]] = i + ;
    printf("%d\n", ans[]);
    for(int i = ; i < ans[]; ++i) printf("%d ", ans[i]);
    printf("%d\n", ans[ans[]]);
    return ;
}