LibreOJ 2058 「TJOI / HEOI2016」求和

设 fn=n∑i=02i⋅i!{ni}。
考虑其组合意义，即 fn 表示 n 个不同的物品放进最多 n 个不同的盒子中，可以有空盒，且每个盒子有黑红两种颜色的方案数。

于是考虑枚举其中一个盒子中物品的个数，易得递推式 fn=2n∑i=1(ni)fn−i。
推一推 fn=2n∑i=1(ni)fn−ifn=2n∑i=1n!i!(n−i)!fn−ifn2⋅n!=n∑i=11i!⋅fn−i(n−i)!

熟悉的卷积形式（其实也不是很熟悉）
于是结合分治 NTT 就有了一个垃圾的 O(nlog2n) 解法……
似乎还有一个 log 的解法和线性的解法……

代码：

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 1 << 18;
const int mod = 998244353;
const int G = 3;
const int Gi = 332748118;
int len,n = 1;
int fpow(int a,int b)
{
    int ret = 1;
    for(;b;b >>= 1)
        (b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
    return ret; 
}
int f[N + 5],omg[N + 5],inv[N + 5];
int fac[N + 5],ifac[N + 5];
int ans;
void ntt(int *a,int *omg,int n,int lg)
{
    for(register int i = 0;i < n;++i)
    {
        int t = 0;
        for(register int j = 0;j < lg;++j)
            if(i & (1 << j))
                t |= (1 << lg - j - 1);
        if(i < t)
            swap(a[i],a[t]);
    }
    for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
        for(register int i = 0;i < n;i += w)
            for(register int j = 0;j < m;++j)
            {
                int t = (long long)omg[n / w * j] * a[i + j + m] % mod;
                a[i + j + m] = (a[i + j] - t + mod) % mod,a[i + j] = (a[i + j] + t) % mod;
            }
}
void mul(int *a,int *b,int n)
{
    int lg = 0;
    for(register int i = 1;i < n;i <<= 1,++lg);
    for(register int i = 0;i < n;++i)
        omg[i] = fpow(G,(mod - 1) / n * i),inv[i] = fpow(Gi,(mod - 1) / n * i);
    ntt(a,omg,n,lg),ntt(b,omg,n,lg);
    for(register int i = 0;i < n;++i)
        a[i] = (long long)a[i] * b[i] % mod;
    ntt(a,inv,n,lg);
    int n_inv = fpow(n,mod - 2);
    for(register int i = 0;i < n;++i)
        a[i] = (long long)a[i] * n_inv % mod;
}
void solve(int l,int r)
{
    if(l == r)
    {
        f[l] = 2LL * f[l] * fac[l] % mod;
        return ;
    }
    int mid = l + r >> 1;
    solve(l,mid);
    static int buf[2][N + 5];
    for(register int i = 0;i <= mid - l;++i)
        buf[0][i] = (long long)f[i + l] * ifac[i + l] % mod;
    for(register int i = mid - l + 1;i <= r - l;++i)
        buf[0][i] = 0;
    for(register int i = 0;i <= r - l;++i)
        buf[1][i] = (long long)ifac[i] % mod;
    mul(buf[0],buf[1],r - l + 1);
    for(register int i = mid + 1;i <= r;++i)
        f[i] = (f[i] + buf[0][i - l]) % mod;
    solve(mid + 1,r);
}
int main()
{
    scanf("%d",&len);
    for(;n < len;n <<= 1);
    fac[0] = 1;
    for(register int i = 1;i < n;++i)
        fac[i] = (long long)fac[i - 1] * i % mod;
    ifac[n - 1] = fpow(fac[n - 1],mod - 2);
    for(register int i = n - 1;i;--i)
        ifac[i - 1] = (long long)ifac[i] * i % mod;
    f[0] = (mod + 1) / 2;
    solve(0,n - 1);
    for(register int i = 0;i <= len;++i)
        ans = (ans + f[i]) % mod;
    printf("%d\n",ans);
}