题面
公式一顿带
化为卷积形式，NTT即可
$O(nlogn)$1
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using namespace std;
typedef long long ll;
const int MAXN = 800010;
const int mod = 167772161, G = 3, Ginv = (167772161 + 1) / 3;
int rev[MAXN];
inline void get_rev(int l) {
for(int i = 1; i < (1 << l); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
}
inline int power(int a, int b) {
ll res = a, ans = 1;
for(; b; b >>= 1, res = res * res % mod) if(b & 1) ans = ans * res % mod;
return ans;
}
int gg[MAXN], ggi[MAXN];
inline void ntt(int *a, int l, int type) {
for(int i = 0; i < (1 << l); i++)
if(rev[i] < i) a[i] ^= a[rev[i]] ^= a[i] ^= a[rev[i]];
for(register int p = 1; p < (1 << l); p <<= 1) {
int wn = (type == 1 ? gg[p] : ggi[p]);
for(register int s = 0; s < (1 << l); s += p << 1) {
int w = 1;
for(register int i = 0; i < p; i++) {
int h1 = a[s + i], h2 = 1ll * w * a[s + p + i] % mod;
a[s + i] = (h1 + h2) % mod;
a[s + p + i] = (h1 - h2 + mod) % mod;
w = 1ll * w * wn % mod;
}
}
}
if(type == -1) {
int inv = power(1 << l, mod - 2);
for(register int i = 0; i < (1 << l); i++) a[i] = 1ll * a[i] * inv % mod;
}
}
inline void add(int *a, int sizea, int *b, int sizeb, int *c) {
for(register int i = 0; i < max(sizea, sizeb); i++)
c[i] = ((i < sizea ? a[i] : 0) + (i < sizeb ? b[i] : 0)) % mod;
}
int f[MAXN], g[MAXN], h[MAXN];
inline void mult(int *a, int sizea, int *b, int sizeb, int *c) {
register int l = 0;
for(; (1 << l) < sizea + sizeb - 1; l++);
get_rev(l);
for(register int i = 0; i < (1 << l); i++) {
f[i] = (i < sizea ? a[i] : 0);
g[i] = (i < sizeb ? b[i] : 0);
}
ntt(f, l, 1);
ntt(g, l, 1);
for(int i = 0; i < (1 << l); i++)
f[i] = 1ll * f[i] * g[i] % mod;
ntt(f, l, -1);
for(register int i = 0; i < sizea + sizeb - 1; i++) c[i] = f[i];
}
int n;
int fac[MAXN], dinv[MAXN], ans[MAXN], finv[MAXN];
int main() {
for(int i = 1; i < MAXN; i <<= 1) gg[i] = power(G, (mod - 1) / 2 / i);
for(int i = 1; i < MAXN; i <<= 1) ggi[i] = power(Ginv, (mod - 1) / 2 / i);
scanf("%d", &n);
fac[0] = 1;
for(int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
for(int i = 0; i <= n; i++) finv[i] = power(i, n);
dinv[n] = power(fac[n], mod - 2);
for(int i = n; i > 0; i--) dinv[i - 1] = 1ll * dinv[i] * i % mod;
for(int i = 0; i <= n; i++) finv[i] = 1ll * finv[i] * dinv[i] % mod;
for(int i = 0; i <= n; i++) dinv[i] = i & 1 ? mod - dinv[i] : dinv[i];
mult(dinv, n + 1, finv, n + 1, ans);
for(int i = 0; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
return 0;
}