bzoj4555

ntt+cdq分治

原来zwh出的cf是斯特林

第二类斯特林数的定义是S(i,j)表示将i个物品分到j个无序集合的方案数，那么这道题中S(i,j)*j!*2^j是指将i个物品分到j个有序集合中并且每个集合可以选或不选的方案数，那么我们改变这个公式，得出

F[i]=∑F[j]*2*C(i,j)，j=0-n,意思是第一个集合选n-j个的方案数，那么这个集合有两种情况选或不选，乘上2，再乘上选出元素的方案数。然后展开组合数，得出F[i]=∑F[j]*2*i!/(i-j)!/j!,移项得出F[i]/i!=∑F[j]/j!*2/(i-j)!

设新的函数G[i]=F[i]/i!,那么G[i]=∑G[j]*2/(i-j)!,后面是卷积形式，用ntt优化，又因为两边都有G，所以我们用cdq分治求和，复杂度nlog^2n

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = ( << ) + , mod = ;
int n;
ll ans;
int rev[N];
ll a[N], b[N], fac[N], facinv[N], inv[N], f[N];
ll power(ll x, ll t)
{
    ll ret = ;
    for(; t; t >>= , x = x * x % mod) if(t & ) ret = ret * x % mod;
    return ret;
}
void ntt(ll *a, int n, int f)
{
    for(int i = ; i < n; ++i) if(i < rev[i]) swap(a[i], a[rev[i]]);
    for(int m = ; m <= n; m <<= )
    {
        int mid = (m >> );
        ll wn = power(, f ==  ? (mod - ) / m : mod -  - (mod - ) / m);
        for(int i = ; i < n; i += m)
        {
            ll w = ;
            for(int j = ; j < mid; ++j)
            {
                ll u = a[i + j], v = a[i + j + mid] * w % mod;
                a[i + j] = (u + v) % mod;
                a[i + j + mid] = (u - v + mod) % mod;
                w = w * wn % mod;
            }
        }
    }
    if(f == -)
    {
        ll inv = power(n, mod - );
        for(int i = ; i < n; ++i) a[i] = a[i] * inv % mod;
    }
}
void cdq(int l, int r)
{
    if(l == r) return;
    int mid = (l + r) >> ;
    cdq(l, mid);
    int lim = r - l + , n = , k = ;
    while(n < lim)
    {
        n <<= ;
        ++k;
    }
    for(int i = ; i < n; ++i) rev[i] = (rev[i >> ] >> ) | ((i & ) << (k - ));
    for(int i = ; i < n; ++i) a[i] = b[i] = ;
    for(int i = l; i <= mid; ++i) a[i - l] = f[i];
    for(int i = ; i < lim; ++i) b[i] = facinv[i];
    ntt(a, n, );
    ntt(b, n, );
    for(int i = ; i < n; ++i) a[i] = a[i] * b[i] % mod;
    ntt(a, n, -);
    for(int i = mid + ; i <= r; ++i) f[i] = (f[i] +  * a[i - l]) % mod;
    cdq(mid + , r);
}
int main()
{
    scanf("%d", &n);
    fac[] = inv[] = facinv[] = ;
    for(int i = ; i <= n; ++i)
    {
        fac[i] = fac[i - ] * i % mod;
        if(i != ) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
        facinv[i] = facinv[i - ] * inv[i] % mod;
    }
    f[] = ;
    cdq(, n);
    for(int i = ; i <= n; ++i) ans = (ans + f[i] * fac[i] % mod) % mod;
    printf("%lld\n", ans);
    return ;
}