Description
计算函数的值
[f(n) = sum limits_{i=0}^{n} sum limits_{j=0}^{i} 2^j imes j! imes S(i,j)
]
Solution
大家好，我是练习推柿子半天的个人练习生(newbielyx)。
[f(n) = sum limits_{i=0}^{n} sum limits_{j=0}^{i} 2^j imes j! imes S(i,j)
]
[= sum limits_{i=0}^{n} sum limits_{j=0}^{i} 2^j imes j! imes frac{1}{j!} imes left (sum limits_{k=0}^{j} (-1)^k C(j,k)(j-k)^i
ight)
]
[= sum limits_{i=0}^{n} sum limits_{j=0}^{i} 2^j sum limits_{k=0}^{j} (-1)^k C(j,k)(j-k)^i
]
[= sum limits_{j=0}^{n}2^j sum limits_{i=j}^{n} sum limits_{k=0}^{j} (-1)^k C(j,k) (j-k)^i
]
[= sum limits_{j=0}^{n} 2^j sum limits_{k=0}^{j} (-1)^k C(j,k) sum limits_{i=0}^{n}(j-k)^i
]
[= sum limits_{j=0}^{n} 2^j sum limits_{k=0}^{j} (-1)^k frac{j!}{k!(j-k)!} sum limits_{i=0}^{n}(j-k)^i
]
[= sum limits_{j=0}^{n} 2^j cdot j! sum limits_{k=0}^{j} frac{(-1)^k}{k!} frac{sum limits_{i=0}^{n}(j-k)^i}{(j-k)!}
]
容易看出上面是一个等比数列，继续推倒
[f(n) = sum limits_{j=0}^{n} 2^j cdot j! sum limits_{k=0}^{j} frac{(-1)^k}{k!} frac{(j-k)^{n+1}-1}{(j-k-1)(j-k)!}
]
[= sum limits_{j=0}^{n} 2^j cdot j! sum limits_{k=0}^{j} frac{(-1)^k}{k!} frac{(j-k)^{n+1}-1}{(j-k)!(j-k-1)}
]
然后你就可以发现，卷起来了！
Code
#include <bits/stdc++.h>
using namespace std;
inline int ty() {
char ch = getchar(); int x = 0, f = 1;
while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef long long ll;
const ll P = 998244353, G = 3, Gx = 332748118;
const int _ = 4e5 + 10;
int N, r[_];
ll fac[_], facinv[_], A[_], B[_];
ll fpow(ll a, ll b) {
ll ret = 1;
for ( ; b; b >>= 1) {
if (b & 1) ret = ret * a % P;
a = a * a % P;
}
return ret;
}
inline void pre() {
fac[0] = fac[1] = 1;
for (int i = 2; i <= N; ++i) fac[i] = fac[i - 1] * i % P;
facinv[0] = 1, facinv[N] = fpow(fac[N], P - 2);
for (int i = N - 1; i >= 1; --i) facinv[i] = facinv[i + 1] * (i + 1) % P;
for (int i = 0; i <= N; ++i) {
A[i] = (i & 1) ? -1 : 1;
A[i] = (A[i] * facinv[i] + P) % P;
}
for (int i = 2; i <= N; ++i) {
B[i] = (fpow(i, N + 1) - 1 + P) % P;
B[i] = B[i] * facinv[i] % P;
B[i] = B[i] * fpow(i - 1, P - 2) % P;
}
B[0] = 1, B[1] = N + 1;
}
void NTT(int lim, ll *a, int op) {
for (int i = 0; i < lim; ++i)
if (i < r[i]) swap(a[i], a[r[i]]);
for (int len = 2; len <= lim; len <<= 1) {
int mid = len >> 1;
ll Wn = fpow(op == 1 ? G : Gx, (P - 1) / len);
for (int i = 0; i < lim; i += len) {
ll w = 1;
for (int j = 0; j < mid; ++j, w = w * Wn % P) {
ll x = a[i + j], y = w * a[i + j + mid] % P;
a[i + j] = (x + y) % P;
a[i + j + mid] = (x - y + P) % P;
}
}
}
}
inline void solve() {
int lim = 1, k = 0;
while (lim <= N + N) lim <<= 1, ++k;
for (int i = 0; i < lim; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (k - 1));
NTT(lim, A, 1);
NTT(lim, B, 1);
for (int i = 0; i < lim; ++i) A[i] = (A[i] * B[i]) % P;
NTT(lim, A, -1);
ll inv = fpow(lim, P - 2);
for (int i = 0; i <= N + N; ++i) A[i] = A[i] * inv % P;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("sum.in", "r", stdin);
freopen("sum.out", "w", stdout);
#endif
N = ty();
pre();
solve();
ll ans = 0, t = 1;
for (int i = 0; i <= N; ++i, t = t * 2 % P) ans = (ans + t * fac[i] % P * A[i] % P) % P;
printf("%lld
", ans);
return 0;
}