bzoj 4555 [Tjoi2016&Heoi2016]求和

求:

$$
\sum_{i=0}^{n}\sum_{j=0}^{i}S_{i}^{j}2^jj!
$$

其中:$1 \le n \le 10^5$,同时 $S_{i}^{j}$ 表示第二类斯特林数

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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N = 5e5 + 10, mod = 998244353;
 5 ll pw(ll a, ll b) {
 6     ll r = 1;
 7     for(a %= mod ; b ; b >>= 1, a = a * a % mod)
 8         if(b & 1)
 9             r = r * a % mod;
10     return r;
11 }
12 const ll G[2] = { 3, pw(3, mod - 2) }; 
13 
14 int n;
15 ll fac[N], invfac[N], a[N], f[N], g[N], h[N];
16 
17 int rev(int x, int n) {
18     int res = 0;
19     for(int i = 0 ; (1 << i) < n ; ++ i)   
20         res = (res << 1) | ((x >> i) & 1);
21     return res;
22 }
23 
24 void ntt(ll *src, int len, int ty) {
25     for(int i = 0 ; i < len ; ++ i) a[rev(i, len)] = src[i];
26     
27     for(int i = 2 ; i <= len ; i <<= 1) {
28         ll wn = pw(G[ty], (mod - 1) / i);
29         for(int j = 0 ; j < len ; j += i) {
30             ll w = 1;
31             for(int k = j ; k < j + i / 2 ; ++ k) {
32                 ll u = a[k], v = w * a[k + i / 2];
33                 a[k] = (u + v) % mod;
34                 a[k + i / 2] = (u - v) % mod;
35                 w = w * wn % mod;
36             }
37         }
38     }
39     
40     ll inv = pw(len, mod - 2);
41     for(int i = 0 ; i < len ; ++ i) {
42         src[i] = a[i];
43         if(ty) {
44             src[i] = src[i] * inv % mod;
45         }
46     }
47 }
48 
49 int main() {
50     scanf("%d", &n);
51     ll ans = 0;
52     
53     int len = 1; while(len <= 2 * (n + 1) + 10) len <<= 1;
54     
55     fac[0] = invfac[0] = 1;
56     for(int i = 1 ; i <= len ; ++ i) fac[i] = fac[i - 1] * i % mod;
57     invfac[len] = pw(fac[len], mod - 2);
58     for(int i = len - 1 ; i >= 1 ; -- i) invfac[i] = invfac[i + 1] * (i + 1) % mod;
59     
60     for(int i = 0 ; i <= n ; ++ i) {
61         f[i] = (i & 1 ? -1 : 1) * invfac[i] % mod;
62         g[i] = i == 0 ? 1 : i == 1 ? n + 1 : (pw(i, n + 1) - 1) % mod * pw(fac[i] * (i - 1) % mod, mod - 2) % mod;
63     }
64     
65     ntt(f, len, 0), ntt(g, len, 0);
66     for(int i = 0 ; i < len ; ++ i) f[i] = f[i] * g[i] % mod;
67     ntt(f, len, 1);
68     
69     for(int i = 0 ; i <= n ; ++ i)
70         ans = (ans + pw(2, i) * fac[i] % mod * f[i] % mod) % mod;
71     printf("%lld\n", (ans % mod + mod) % mod);
72 } 
bzoj 4555 [Tjoi2016&Heoi2016]求和

posted @ 2018-12-12 16:04 KingSann 阅读(...) 评论(...) 编辑 收藏