【LOJ】#150. 挑战多项式

原题链接

多项式全家桶！快乐！（好像少个除法，不过有除法好像不太快乐）

(说真的这是我第一次写exp和开根。。。水平不行。。

从最基础要实现的操作开始吧。。

多项式取模$x^n$

这个。。很简单了，把大于等于n指数的系数全改成0就好了

多项式加减法

按位加/减，复杂度$O(n)$

多项式求导数

这个的话，选修2-1了解一下？

$(ax^{n})’ = anx^{n - 1}$

多项式求积分

同上

我们只要求一个多项式的导数是该多项式即可

列个方程可以发现

$\int ax^{n} = \frac{a}{n + 1}x^{n + 1}$

多项式乘法

FFT了解一下？？不过这题是NTT，复杂度$O(n \log n)$

多项式求逆元

也就是$A(x)B(x) \equiv 1 \pmod{x^{2^{t}}}$

我们倍增一下，当$t = 0$的时候直接求$\frac{1}{A(0)}$

如果不是的话有

$B(x) \equiv G(x) \pmod{x^{2^{t}}}$

我们把$G(x)$移过去

$B(x) - G(x) \equiv 0 \pmod {x^{2^{t}}}$

两边平方一下

$(B(x) - G(x))^{2} \equiv 0 \pmod {x^{2^{t + 1}}}$

$B(x)^{2} - 2B(x)G(x) + G(x)^{2} \equiv 0 \pmod {x^{2^{t + 1}}}$

$B(x)^{2} \equiv 2B(x)G(x) - G(x)^{2} \pmod {x^{2^{t + 1}}}$

两边同乘一个$A(x)$

$B(x) \equiv 2G(x) - A(x)G(x)^{2} \pmod {x^{2^{t + 1}}}$

就可以算了

多项式开根

和上面的构造很类似！

$B(x)^2 = A(x) \pmod{x^{2^{t}}}$

当$t = 0$，我们求$\sqrt{A(0)}$，这个可以写cipolla

然后有

$B(x) \equiv G(x) \pmod{x^{2^{t}}}$

$B(x) - G(x) \equiv 0 \pmod{x^{2^{t}}}$

再平方

$B(x)^{2} - 2B(x)G(x) + G(x)^2 \equiv 0 \pmod{x^{2^{t + 1}}}$

然后把$B(x)^2$换成$A(x)$

$A(x)- 2B(x)G(x) + G(x)^2 \equiv 0 \pmod{x^{2^{t + 1}}}$

整理一下

$B(x) \equiv \frac{A(x) + G(x)^2}{2G(x)} \pmod{x^{2^{t + 1}}}$

多项式求ln

$\ln F(x) = G(x)$

两边分别求导

$\frac{F'(x)}{F(x)} = G'(x)$

然后再两边一起积分

$\int \frac{F'(x)}{F(x)} = G(x)$

所以多项式求ln只要求个导数，求个逆元，求个积分就好了

多项式求exp

根据牛顿迭代我们有。。

如果有个函数

$B(A_t(x)) \equiv 0 \pmod{x^{2^{t}}}$

$A_{t + 1}(x) = A_{t}(x) - \frac{B(A_t(x))}{B'(A_t(x))}$

注意下面求导的主元是$A_t(x)$而不是$x$

这个可以用泰勒展开展开证一下。。

$B(A_{t + 1}(x)) \equiv 0 \pmod{x^{2^{t + 1}}}$

$0 = B(A_{t + 1}(x)) = B(A_t(x)) + \frac{B'(A_t(x))(A_{t + 1}(x) - A_{t}(x))}{1!} + \frac{B''(A_t(x))(A_{t + 1}(x) - A_t(x))^2}{2!}\cdots$

自第三项及以后，最低项就是$x^{2^{t + 1}}$

所以可以化简成

$0 = B(A_t(x)) + B'(A_{t}(x))(A_{t + 1}(x) - A_t(x))$

然后就可以得到

$A_{t + 1}(x) = A_t(x) - \frac{B(A_t(x))}{B'(A_{t}(x))}$

然后……

$e^{F(x)} \equiv G_{t}(x) \pmod{x^{2^{t}}}$

然后同时取$\ln$

$F(x) - \ln G_t(x) \equiv 0 \pmod {x^{2^{t}}}$

然后就是刚刚那个式子咯。。

$G_{t + 1}(x) \equiv G_{t}(x) - \frac{F(x) - \ln(G(x))}{-\frac{1}{G(x)}}\pmod{x^{2^{t + 1}}}$

还是刚刚那句话。。主元是$G_t(x)$不是$x$，所以$F(x)$相当于一个常数项。。

$G_{t + 1}(x) \equiv G_{t}(x) + G_t(x)(F(x) - \ln(G(x))) \pmod {x^{2^{t + 1}}}$

然后就可以做了

多项式快速幂

$A^{k}(x) = e^{k \ln A(x)}$

写的话有个小细节。。就是注意运算的时候指数取模不是对N取模，是对$2^t$取模。。

#include <bits/stdc++.h>

#define fi first

#define se second

#define pii pair<int,int>

#define mp make_pair

#define pb push_back

#define space putchar(' ')

#define enter putchar('\n')

#define MAXN 20000005

#define eps 1e-10

//#define ivorysi

using namespace std;

typedef long long int64;

typedef unsigned int u32;

typedef double db;

template<class T>

void read(T &res) {

	res = 0;T f = 1;char c = getchar();

	while(c < '0' || c > '9') {

		if(c == '-') f = -1;

		c = getchar();

	}

	while(c >= '0' && c <= '9') {

		res = res * 10 + c - '0';

		c = getchar();

	}

	res *= f;

}

template<class T>

void out(T x) {

	if(x < 0) {x = -x;putchar('-');}

	if(x >= 10) {

		out(x / 10);

	}

	putchar('0' + x % 10);

}

const int MOD = 998244353,MAXL = (1 << 21);

int W[MAXL + 5],N,K,Inv[MAXL + 5];

int inc(int a,int b) {

	return a + b >= MOD ? a + b - MOD : a + b;

}

int mul(int a,int b) {

	return 1LL * a * b % MOD;

}

int fpow(int x,int c) {

	int res = 1,t = x;

	while(c) {

		if(c & 1) res = mul(res,t);

		t = mul(t,t);

		c >>= 1;

	}

	return res;

}

namespace Cipolla {

    int a,s;

    struct Complex {

        int r,i;

        Complex(int _x = 0,int _y = 0) {r = _x;i = _y;}

        friend Complex operator + (const Complex &a,const Complex &b) {

            return Complex(inc(a.r,b.r),inc(a.i,b.i));

        }

        friend Complex operator - (const Complex &a,const Complex &b) {

            return Complex(inc(a.r,MOD - b.r),inc(a.i,MOD - b.i));

        }

        friend Complex operator * (const Complex &a,const Complex &b) {

            return Complex(inc(mul(a.r,b.r),mul(mul(a.i,b.i),s)),inc(mul(a.r,b.i),mul(a.i,b.r)));

        }

    };

    u32 Rand() {

        static u32 x = 20020421;

        return x += x << 2 | 1;

    }

    Complex Fpow(Complex x,int c) {

        Complex res(1,0),t = x;

        while(c) {

            if(c & 1) res = res * t;

            t = t * t;

            c >>= 1;

        }

        return res;

    }

    int Sqrt(int N) {

        while(1) {

            a = Rand() % MOD;

            s = inc(mul(a,a),MOD - N);

            if(fpow(s,(MOD - 1) / 2) == MOD - 1) break;

        }

        Complex res(a,1);

        res = Fpow(res,(MOD + 1) / 2);

        if(res.r > MOD - res.r) res = MOD - res.r;

        return res.r;

     }

};

struct Poly {

	vector<int> v;

	Poly() {v.clear();}

	Poly(int a) {v.clear();v.pb(a);}

	Poly limit(int len = N) {

		if(!v.size()) v.pb(0);

		if(v.size() > len) v.resize(len);

		int t = v.size() - 1;

		while(t && v[t] == 0) {v.pop_back();--t;}

		return *this;

	}

	friend Poly operator + (Poly a,Poly b) {

		int h = max(a.v.size(),b.v.size());

		a.v.resize(h);b.v.resize(h);

		Poly c;

		for(int i = 0 ; i < h ; ++i) {

			c.v.pb(inc(a.v[i],b.v[i]));

		}

		return c;

	}

	friend Poly operator - (Poly a,Poly b) {

		int h = max(a.v.size(),b.v.size());

		a.v.resize(h),b.v.resize(h);

		Poly c;

		for(int i = 0 ; i < h ; ++i) {

			c.v.pb(inc(a.v[i],MOD - b.v[i]));

		}

		return c;

	}

	friend void NTT(Poly &a,int L,int on) {

		a.v.resize(L);

		for(int i = 1 ,j = L >> 1 ; i < L - 1 ; ++i) {

			if(i < j) swap(a.v[i],a.v[j]);

			int k = L >> 1;

			while(j >= k) {

				j -= k;

				k >>= 1;

			}

			j += k;

		}

		for(int h = 2 ; h <= L ; h <<= 1) {

			int wn = W[(MAXL + MAXL / h * on) % MAXL];

			for(int k = 0 ; k < L ; k += h) {

				int w = 1;

				for(int j = k ; j < k + h / 2 ; ++j) {

					int u = a.v[j],t = mul(a.v[j + h / 2],w);

					a.v[j] = inc(u,t);

					a.v[j + h / 2] = inc(u,MOD - t);

					w = mul(w,wn);

				}

			}

		}

		if(on == -1) {

			int InvL = fpow(L,MOD - 2);

			for(int i = 0 ; i < L ; ++i) a.v[i] = mul(a.v[i],InvL);

		}

	}

	friend Poly operator * (const int &a,const Poly &b) {

		Poly c;int l = b.v.size();

		for(int i = 0 ; i < l ; ++i) c.v.pb(mul(a,b.v[i]));

		return c;

	}

	friend Poly operator * (Poly a,Poly b) {

		Poly c;

		int t = a.v.size() + b.v.size(),l = 1;

		while(l <= t) l <<= 1;

		a.v.resize(l);b.v.resize(l);

		NTT(a,l,1);NTT(b,l,1);

		c.v.resize(l);

		for(int i = 0 ; i < l ; ++i) c.v[i] = mul(a.v[i],b.v[i]);

		NTT(c,l,-1);

		return c.limit();

	}

	friend Poly inverse(Poly a,int t) {

		if(t == 1) return Poly(fpow(a.v[0],MOD - 2));

		a.limit(t);

		Poly b = inverse(a,t >> 1);

		int l = b.v.size() + b.v.size() + a.v.size();

		int len = 1;

		while(len <= l) len <<= 1;

		NTT(b,len,1);NTT(a,len,1);

		Poly c;c.v.resize(len);

		for(int i = 0 ; i < len ; ++i) c.v[i] = inc(mul(2,b.v[i]),MOD - mul(mul(b.v[i],b.v[i]),a.v[i]));

		NTT(c,len,-1);

		return c.limit(t);

	}

	friend Poly Inverse(const Poly &a,int t = N) {

		int l = 1;

		while(l < t) l <<= 1;

		return inverse(a,l).limit(t);

	}

	friend Poly sub_sqrt(Poly a,int t) {

		if(t == 1) return Poly(Cipolla::Sqrt(a.v[0]));

		a.limit(t);

		Poly b = sub_sqrt(a,t >> 1);

		Poly inv = 2 * b;inv.v.resize(t);inv = Inverse(inv,t);

		int l = inv.v.size() + max(b.v.size() + b.v.size(),a.v.size()),len = 1;

		while(len <= l) len <<= 1;

		NTT(inv,len,1);NTT(b,len,1);NTT(a,len,1);

		Poly c;c.v.resize(len);

		for(int i = 0 ; i < len ; ++i) {

			c.v[i] = mul(inc(a.v[i],mul(b.v[i],b.v[i])),inv.v[i]);

		}

        NTT(c,len,-1);

		return c.limit(t);

	}

	friend Poly Sqrt(const Poly &a,int t = N) {

		int l = 1;

		while(l < t) l <<= 1;

		return sub_sqrt(a,l).limit(t);

	}

	friend Poly Derivative(const Poly &a) {

		Poly c;c.v.resize(a.v.size() - 1);

		for(int i = 0 ; i < a.v.size() - 1 ; ++i) {

			c.v[i] = mul(a.v[i + 1],i + 1);

		}

		return c;

	}

	friend Poly Integral(const Poly &a) {

		Poly c;c.v.resize(a.v.size() + 1);

		for(int i = 1 ; i <= a.v.size() ; ++i) {

			c.v[i] = mul(a.v[i - 1],Inv[i]);

		}

		return c;

	}

	friend Poly ln(Poly a,int t = N) {

		Poly c = Inverse(a,t),d = Derivative(a);

		int l = c.v.size() + d.v.size(),len = 1;

		while(len <= l) len <<= 1;

		NTT(c,len,1);NTT(d,len,1);

		Poly res;res.v.resize(len);

		for(int i = 0 ; i < len ; ++i) res.v[i] = mul(c.v[i],d.v[i]);

		NTT(res,len,-1);

		return Integral(res).limit(t);

	}

	friend Poly sub_exp(Poly a,int t) {

		if(t == 1) {return Poly(1);}

		a.limit(t);

		Poly b = sub_exp(a,t >> 1);

		Poly lnb = ln(b,t);

		int l = b.v.size() + max(lnb.v.size(),a.v.size()),len = 1;

		while(len <= l) len <<= 1;

		NTT(lnb,len,1);NTT(b,len,1);NTT(a,len,1);

		Poly c;c.v.resize(len);

		for(int i = 0 ; i < len ; ++i) {

			c.v[i] = inc(b.v[i],mul(b.v[i],inc(a.v[i],MOD - lnb.v[i])));

		}

		NTT(c,len,-1);

		return c.limit(t);

	}

	friend Poly Exp(Poly a) {

		int l = 1;

		while(l < a.v.size()) l <<= 1;

		return sub_exp(a,l).limit();

	}

	friend Poly operator ^ (const Poly &a,const int &k) {

		return Exp(k * ln(a)).limit();

	}

	friend void Print(const Poly &a) {

		for(int i = 0 ; i < a.v.size() ; ++i) {

			out(a.v[i]);space;

		}

		enter;

	}

}F,G;

void Solve() {

	read(N);read(K);

	int a;

	for(int i = 0 ; i <= N ; ++i) {

		read(a);

		F.v.pb(a);

	}

	Inv[1] = 1;W[0] = 1;W[1] = fpow(3,(MOD - 1) / MAXL);

	for(int i = 2 ; i < MAXL ; ++i) {

		W[i] = mul(W[i - 1],W[1]);

		Inv[i] = mul(Inv[MOD % i],MOD - MOD / i);

	}

	int tip = N;

	N++;

	G = Derivative((Poly(1) + ln(Poly(2) + F - Poly(F.v[0]) - Exp(Integral(Inverse(Sqrt(F))))))^K);

	G.v.resize(tip);

	Print(G);

	enter;

}

int main() {

#ifdef ivorysi

	freopen("f1.in","r",stdin);

#endif

	Solve();

}