题意：
求
[f(n)=sum_{i=0}^nsum_{j=0}^iegin{Bmatrix}
i \ j
end{Bmatrix}2^jj!
]
思路：
直接将第二类斯特林数展开有：
[egin{aligned}
f(n)=&sum_{i=0}^nsum_{j=0}^n2^jsum_{k=0}^{j}(-1)^k{jchoose k}(j-k)^{i}\
=&sum_{i=0}^nsum_{j=0}^n2^jj!sum_{k=0}^jfrac{(-1)^k}{k!}frac{(j-k)^i}{(j-k)!}\
=&sum_{j=0}^n2^jj!sum_{k=0}^jfrac{(-1)^k}{k!}frac{sum_{i=0}^n(j-k)^i}{(j-k)!}
end{aligned}
]
观察到后半部分为一个卷积的形式，我们令(displaystyle a_i=frac{(-1)^i}{i!},b_i=frac{sum_{j=0}^ni^j}{i!})，其中(sum_{j=0}^n i^j)为等比数列求和的形式。那么直接将这两个作为系数卷一卷即可。
还有一种做法为展开递推式的做法，忘了怎么做了。。明天来补。
upd:做法如下：
[egin{aligned}
f(n)=sum_{j=0}^n2^jj!sum_{i=j}^negin{Bmatrix}
i \ j
end{Bmatrix}
end{aligned}
]
令(displaystyle F(j)=sum_{i=j}^negin{Bmatrix} i \ j end{Bmatrix})，之后将第二类斯特林数用递推式展开：
[egin{aligned}
F(j)=&sum_{i=j}^negin{Bmatrix}
i \ j
end{Bmatrix}\
=&sum_{i=j}^negin{Bmatrix}
i - 1 \ j - 1
end{Bmatrix}+sum_{i=j}^njegin{Bmatrix}
i - 1 \ j
end{Bmatrix}\
=&F(j-1)-egin{Bmatrix}
n \ j - 1
end{Bmatrix}+jF(j)-jegin{Bmatrix}
n \ j
end{Bmatrix}\
=&F(j - 1)+jF(j)-egin{Bmatrix}
n + 1 \ j
end{Bmatrix}
end{aligned}
]
所以最后有：
[F(j)=frac{F(j-1)-egin{Bmatrix}
n + 1 \ j
end{Bmatrix}}{1-j}
]
那么预处理出(n+1)这一行的斯特林数，然后直接用递推式算出(F)即可。
代码是第二种方法：
/*
* Author: heyuhhh
* Created Time: 2019/12/12 11:16:54
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <iomanip>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << '
'; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
#define dbg(...)
#endif
void pt() {std::cout << '
'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 4e5 + 5, M = 2e6 + 5, P = 998244353, G = 3, Gi = 332748118, MOD = 998244353;
int n, m, lim = 1, L, r[N];
ll a[N], b[N];//注意空间要开四倍
ll qpow(ll a, ll k) {
ll ans = 1;
while(k) {
if(k & 1) ans = (ans * a ) % P;
a = (a * a) % P;
k >>= 1;
}
return ans;
}
void NTT(ll *A, int type) {
for(int i = 0; i < lim; i++)
if(i < r[i]) swap(A[i], A[r[i]]);
for(int mid = 1; mid < lim; mid <<= 1) {
ll Wn = qpow( type == 1 ? G : Gi , (P - 1) / (mid << 1)); //Wn = g ^ ((p - 1) / n) (mod p)
for(int j = 0; j < lim; j += (mid << 1)) {
ll w = 1;
for(int k = 0; k < mid; k++, w = (w * Wn) % P) {
int x = A[j + k], y = w * A[j + k + mid] % P;
A[j + k] = (x + y) % P,
A[j + k + mid] = (x - y + P) % P;
}
}
}
}
void solve(ll *a, ll *b) {
while(lim <= n + m) lim <<= 1, L++;
for(int i = 0; i < lim; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
for(int i = m + 1; i < lim; i++) a[i] = 0; //a,b need init
for(int i = m + 1; i < lim; i++) b[i] = 0;
NTT(a, 1); NTT(b, 1);
for(int i = 0; i < lim; i++) a[i] = (a[i] * b[i]) % P;
NTT(a, -1);
ll inv = qpow(lim, P - 2);
for(int i = 0; i < lim; i++) a[i] = a[i] * inv % P;
}
int fac[N], inv[N], f[N], two[N];
void init() {
fac[0] = 1;
for(int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
inv[N - 1] = qpow(fac[N - 1], MOD - 2);
for(int i = N - 2; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
for(int i = 0; i <= m; i++) {
a[i] = (i & 1) ? MOD - inv[i] : inv[i];
b[i] = 1ll * qpow(i, n + 1) * inv[i] % MOD;
}
solve(a, b);
f[1] = n;
for(int i = 2; i <= n; i++) f[i] = 1ll * (a[i] - f[i - 1] + MOD) % MOD * qpow(i - 1, MOD - 2) % MOD;
two[0] = 1;
for(int i = 1; i <= n; i++) two[i] = 1ll * two[i - 1] * 2 % MOD;
}
void run(){
cin >> n; m = n + 1;
init();
int ans = 1;
for(int i = 1; i <= n; i++) {
ans = (ans + 1ll * two[i] * fac[i] % MOD * f[i] % MOD) % MOD;
}
cout << ans << '
';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
run();
return 0;
}