[TJOI 2016&HEOI 2016]求和

Description

题库链接

求

$$f(n)=\sum_{i=0}^n\sum_{j=0}^i S(i,j)\times 2^j \times (j!)$$

$S(i, j)$ 表示第二类斯特林数。对 $998244353$ 取模。

$1\leq n\leq 100000$

Solution

由于 $S(i,j)=0,i\leq j$ ，我们可以把式子改写成

$$f(n)=\sum_{i=0}^n\sum_{j=0}^n S(i,j)\times 2^j \times (j!)$$

那么

$$f(n)=\sum_{j=0}^n 2^j \times (j!)\times\sum_{i=0}^n S(i,j)$$

把 $S(i, j)$ 的通项公式代入

$$\begin{aligned}f(n)&=\sum_{j=0}^n 2^j \times (j!)\times\sum_{i=0}^n \sum_{k=0}^j\frac{(-1)^k}{k!}\frac{(j-k)^i}{(j-k)!}\&=\sum_{j=0}^n 2^j \times (j!)\times\sum_{k=0}^j\frac{(-1)^k}{k!}\frac{\sum\limits_{i=0}^n(j-k)^i}{(j-k)!}\end{aligned}$$

记

$$\begin{aligned}A(x)&=\sum_{i=0}^\infty \frac{(-1)^i}{i!}x^i\B(x)&=\sum_{i=0}^\infty\frac{\sum\limits_{k=0}^ni^k}{i!}x^i\end{aligned}$$

那么

$$f(n)=\sum_{j=0}^n 2^j\times(j!)\times(A\otimes B)(j)$$

$\text{NTT}$ 优化即可。

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 100000*4, yzh = 998244353;

int n, inv[N+5], a[N+5], b[N+5], len, L, R[N+5];

int quick_pow(int a, int b) {
    int ans = 1;
    while (b) {
        if (b&1) ans = 1ll*ans*a%yzh;
        b >>= 1, a = 1ll*a*a%yzh;
    }
    return ans;
}
void NTT(int *A, int o) {
    for (int i = 0; i < len; i++) if (i < R[i]) swap(A[i], A[R[i]]);
    for (int i = 1; i < len; i <<= 1) {
        int gn = quick_pow(3, (yzh-1)/(i<<1)), x, y;
        if (o == -1) gn = quick_pow(gn, yzh-2);
        for (int j = 0; j < len; j += (i<<1)) {
            int g = 1;
            for (int k = 0; k < i; k++, g = 1ll*g*gn%yzh) {
                x = A[j+k], y = 1ll*g*A[j+k+i]%yzh;
                A[j+k] = (x+y)%yzh, A[j+k+i] = (x-y)%yzh;
            }
        }
    }
}
void work() {
    scanf("%d", &n); inv[0] = inv[1] = 1;
    for (int i = 2; i <= n; i++) inv[i] = -1ll*yzh/i*inv[yzh%i]%yzh;
    for (int i = 1; i <= n; i++) inv[i] = 1ll*inv[i]*inv[i-1]%yzh;
    for (int i = 0; i <= n; i++)
        if (i&1) a[i] = -inv[i]; else a[i] = inv[i];
    b[0] = 1; b[1] = n+1;
    for (int i = 2; i <= n; i++)
        b[i] = 1ll*inv[i]*(quick_pow(i, n+1)-1)%yzh*quick_pow(i-1, yzh-2)%yzh;
    for (len = 1; len <= (n<<1); len <<= 1) L++;
    for (int i = 0; i < len; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
    NTT(a, 1), NTT(b, 1);
    for (int i = 0; i < len; i++) a[i] = 1ll*a[i]*b[i]%yzh;
    NTT(a, -1);
    for (int i = 0, inv = quick_pow(len, yzh-2); i < len; i++)
        a[i] = 1ll*a[i]*inv%yzh;
    int ans = 0;
    for (int i = 0, ad = 1; i <= n; i++, ad = 2ll*ad%yzh*i%yzh)
        (ans += 1ll*a[i]*ad%yzh) %= yzh;
    printf("%d\n", (ans+yzh)%yzh);
}
int main() {work(); return 0; }