题目
在2016年，佳媛姐姐刚刚学习了第二类斯特林数，非常开心。
现在他想计算这样一个函数的值:
S(i, j)表示第二类斯特林数，递推公式为:
S(i, j) = j ? S(i ? 1, j) + S(i ? 1, j ? 1), 1 <= j <= i ? 1。
边界条件为：S(i, i) = 1(0 <= i), S(i, 0) = 0(1 <= i)
你能帮帮他吗?
输入格式
输入只有一个正整数
输出格式
输出f(n)。由于结果会很大，输出f(n)对998244353(7 × 17 × 223 + 1)取模的结果即可。1 ≤ n ≤ 100000
输入样例
3
输出样例
87
题解
当第二类斯特林数\(j > i\)时值为\(0\)
所以我们实际求：
\[
\begin{aligned}
ans &= \sum\limits_{i = 0}^{n} \sum\limits_{j = 0}^{n} \begin{Bmatrix} i \\ j \end{Bmatrix} 2^{j}j! \&= \sum\limits_{i = 0}^{n} \sum\limits_{j = 0}^{n} 2^{j}j! \frac{1}{j!} \sum\limits_{k = 0}^{j} (-1)^{k}{j \choose k}(j - k)^{i} \&= \sum\limits_{i = 0}^{n} \sum\limits_{j = 0}^{n} 2^{j}j! \sum\limits_{k = 0}^{j} \frac{(-1)^{k}}{k!} * \frac{(j - k)^{i}}{(j - k)!} \&= \sum\limits_{j = 0}^{n} 2^{j}j! \sum\limits_{k = 0}^{j} \frac{(-1)^{k}}{k!} * \frac{\sum\limits_{i = 0}^{n} (j - k)^{i}}{(j - k)!} \\end{aligned}
\]
NTT即可
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
const int G = 3,P = 998244353;
int fac[maxn],fv[maxn],inv[maxn],bin[maxn],g[maxn];
int L,R[maxn],A[maxn],B[maxn],n,m,N;
inline int qpow(int a,int b){
int ans = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) ans = 1ll * ans * a % P;
return ans;
}
void init(){
fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
bin[0] = 1; bin[1] = 2;
g[0] = 1; g[1] = N + 1;
for (int i = 2; i <= N; i++){
fac[i] = 1ll * fac[i - 1] * i % P;
inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
fv[i] = 1ll * fv[i - 1] * inv[i] % P;
bin[i] = 2ll * bin[i - 1] % P;
g[i] = 1ll * (1ll * qpow(i,N + 1) - 1 + P) % P * inv[i - 1] % P;
}
}
void NTT(int* a,int f){
for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k]; y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P; a[j + k + i] = ((x - y) % P + P) % P;
}
}
}
if (f == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int main(){
N = read();
init();
for (int i = 0; i <= N; i++){
A[i] = ((i & 1) ? -1ll : 1ll) * fv[i] % P;
B[i] = 1ll * g[i] * fv[i] % P;
}
m = N + N; L = 0;
for (n = 1; n <= m; n <<= 1) L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
NTT(A,1); NTT(B,1);
for (int i = 0; i < n; i++) A[i] = 1ll * A[i] % P * B[i] % P;
NTT(A,-1);
int ans = 0;
for (int i = 0; i <= N; i++)
ans = (ans + 1ll * bin[i] * fac[i] % P * A[i] % P) % P;
printf("%d\n",(ans % P + P) % P);
return 0;
}原文地址：https://www.cnblogs.com/Mychael/p/8975630.html