[TJOI 2016&HEOI 2016]求和

Description

题库链接

求

[f(n)=sum_{i=0}^nsum_{j=0}^i S(i,j) imes 2^j imes (j!)]

(S(i, j)) 表示第二类斯特林数。对 (998244353) 取模。

(1leq nleq 100000)

Solution

由于 (S(i,j)=0,ileq j) ，我们可以把式子改写成

[f(n)=sum_{i=0}^nsum_{j=0}^n S(i,j) imes 2^j imes (j!)]

那么

[f(n)=sum_{j=0}^n 2^j imes (j!) imessum_{i=0}^n S(i,j)]

把 (S(i, j)) 的通项公式代入

[egin{aligned}f(n)&=sum_{j=0}^n 2^j imes (j!) imessum_{i=0}^n sum_{k=0}^jfrac{(-1)^k}{k!}frac{(j-k)^i}{(j-k)!}\&=sum_{j=0}^n 2^j imes (j!) imessum_{k=0}^jfrac{(-1)^k}{k!}frac{sumlimits_{i=0}^n(j-k)^i}{(j-k)!}end{aligned}]

记

[egin{aligned}A(x)&=sum_{i=0}^infty frac{(-1)^i}{i!}x^i\B(x)&=sum_{i=0}^inftyfrac{sumlimits_{k=0}^ni^k}{i!}x^iend{aligned}]

那么

[f(n)=sum_{j=0}^n 2^j imes(j!) imes(Aotimes B)(j)]

( ext{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
", (ans+yzh)%yzh);
}
int main() {work(); return 0; }