【COGS2187】帕秋莉的超级多项式

Description

求：

Solution

直接模拟即可，具体操作可参考之前的博客。

多项式幂次还没有写过，其实很简单，$F^k=e^{k\ln F}$即可，注意两点：

• 如果$F$有常数项不是$1$或者没有常数项要特殊处理
• 泰勒展开后的$k$并没有出现在指数上，所以直接对模数取模而不是模数-1。

Code

/************************************************* Au: Hany01* Date: Jul 30th, 3018* Prob: 帕秋莉的超级多项式* Email: hany01@foxmail.com* Inst: Yali High School
************************************************/#include<bits/stdc++.h>using namespace std;typedef long long LL;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaiatemplate <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }inline int read() {static int _, __; static char c_;for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);return _ * __;
}const int MOD = 998244353, g0 = 3, maxn = 1e5 + 5;int rev[maxn << 2], powg[maxn << 2], ipowg[maxn << 2], ig0, invn;inline int ad(int x, int y) { if ((x += y) >= MOD) return x - MOD; return x; }inline LL Pow(LL a, LL b) {static LL Ans;for (Ans = 1; b; b >>= 1, (a *= a) %= MOD) if (b & 1) (Ans *= a) %= MOD;return Ans;
}inline void Init(int N) {ig0 = Pow(g0, MOD - 2);for (register int i = 1; i <= N; i <<= 1)powg[i] = Pow(g0, (MOD - 1) / i), ipowg[i] = Pow(ig0, (MOD - 1) / i);
}inline void pre(int n) {static int N, cnt;for (N = 1, cnt = 0; N < n; N <<= 1, ++ cnt);rep(i, n) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));invn = Pow(n, MOD - 2);
}inline void NTT(int* a, int n, int ty) {rep(i, n) if (i < rev[i]) swap(a[i], a[rev[i]]);for (register int i = 2, p = 1; i <= n; p = i, i <<= 1) {register int w0 = ty > 0 ? powg[i] : ipowg[i];for (register int j = 0; j < n; j += i) {register int w = 1;rep(k, p) {register int x = a[j + k], y = (LL)a[j + k + p] * w % MOD;a[j + k] = ad(x, y), a[j + k + p] = ad(x, MOD - y);w = (LL)w * w0 % MOD;}}}if (ty < 1) rep(i, n) a[i] = (LL)a[i] * invn % MOD;
}inline void Mult(int* a, int* b, int* c, int n) {static int A[maxn << 2], B[maxn << 2];rep(i, n) A[i] = a[i], B[i] = b[i];pre(n), NTT(A, n, 1), NTT(B, n, 1);rep(i, n) A[i] = (LL)A[i] * B[i] % MOD;NTT(A, n, -1);rep(i, n) c[i] = A[i];
}namespace Inv {static int A[maxn << 2], B[maxn << 2], a[maxn << 2];void Inv_(int* b, int n) {if (n == 1) { b[0] = Pow(a[0], MOD - 2); return; }Inv_(b, n >> 1);rep(i, n) A[i] = a[i], B[i] = b[i];For(i, n, n << 1) A[i] = B[i] = 0;pre(n << 1), NTT(A, n << 1, 1), NTT(B, n << 1, 1);rep(i, n << 1) A[i] = (LL)A[i] * B[i] % MOD * B[i] % MOD;NTT(A, n << 1, -1);rep(i, n) b[i] = ad(ad(b[i], b[i]), MOD - A[i]);}inline void Inv(int* x, int* y, int n) {rep(i, n) a[i] = x[i], y[i] = 0;Inv_(y, n);}
}inline void Int(int* a, int* b, int n) {Fordown(i, n - 1, 0) b[i + 1] = Pow(i + 1, MOD - 2) * a[i] % MOD;//downto!!!b[0] = 0;
}inline void Der(int *a, int* b, int n) {rep(i, n - 1) b[i] = (LL)a[i + 1] * (i + 1) % MOD;
}inline void Ln(int* a, int* b, int n) {static int A[maxn << 2], B[maxn << 2];rep(i, n) A[i] = B[i] = a[i];For(i, n, n << 1) A[i] = B[i] = 0;Inv:: Inv(A, A, n), Der(B, B, n << 1);Mult(A, B, A, n << 1), Int(A, A, n << 1);rep(i, n) b[i] = A[i];
}namespace Exp {static int a[maxn << 2], A[maxn << 2];void Exp_(int* b, int n) {if (n == 1) { b[0] = 1; return; }Exp_(b, n >> 1), Ln(b, A, n);rep(i, n) A[i] = ad(a[i], MOD - A[i]);For(i, n, n << 1) A[i] = 0, b[i] = 0;++ A[0], Mult(b, A, b, n << 1);}void Exp(int* x, int* y, int n) {rep(i, n) a[i] = x[i], y[i] = 0;Exp_(y, n);}
}namespace Sqrt
{static int A[maxn << 2], B[maxn << 2], a[maxn << 2];void Sqrt_(int* b, int n) {if (n == 1) { b[0] = sqrt(a[0]); return; }Sqrt_(b, n >> 1);rep(i, n) A[i] = b[i];For(i, n, n << 1) A[i] = 0;pre(n << 1), NTT(A, n << 1, 1);rep(i, n << 1) A[i] = (LL)A[i] * A[i] % MOD;NTT(A, n << 1, -1);rep(i, n) A[i] = ad(A[i], a[i]), B[i] = ad(b[i], b[i]);Inv:: Inv(B, B, n);pre(n << 1), NTT(A, n << 1, 1), NTT(B, n << 1, 1);rep(i, n << 1) A[i] = (LL)A[i] * B[i] % MOD;NTT(A, n << 1, -1); rep(i, n) b[i] = A[i];}void Sqrt(int* x, int* y, int n) {rep(i, n) a[i] = x[i], y[i] = 0;Sqrt_(y, n);}
}inline void PPow(int* a, int k, int* b, int n) {Ln(a, b, n);rep(i, n) b[i] = (LL)b[i] * k % MOD;Exp:: Exp(b, b, n);
}int main()
{freopen("polynomial.in", "r", stdin);freopen("polynomial.out", "w", stdout);static int n, k, f[maxn << 2], N;for (n = read(), k = read(), N = 1; N < n; N <<= 1);Init(N << 1);rep(i, n) f[i] = read();Sqrt:: Sqrt(f, f, N);Inv:: Inv(f, f, N);Int(f, f, N);Exp:: Exp(f, f, N);Inv:: Inv(f, f, N), ++ f[0];Ln(f, f, N), ++ f[0];PPow(f, k, f, N);Der(f, f, N);rep(i, n - 1) printf("%d ", f[i]);puts("0");return 0;
}