P4726 【模板】多项式指数函数

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给出 (n-1) 次多项式 (A(x))，求一个 (mod{:x^n}) 下的多项式 (B(x))，满足 (B(x) equiv e^{A(x)})

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第一行一个整数 (n).

下一行有 (n) 个整数，依次表示多项式的系数 (a_0, a_1, cdots, a_{n-1})

保证 (a_0 = 0).

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输出 (n) 个整数，表示答案多项式中的系数 (a_0, a_1, cdots, a_{n-1}).

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6
0 927384623 817976920 427326948 149643566 610586717

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1 927384623 878326372 3882 273455637 998233543

(color{#0066ff}{数据范围与提示})

对于 (100\%)的数据，(n le 10^5).

(color{#0066ff}{ 题解 })

牛顿迭代搞一下

(F(x)=e^{A(x)})

(ln F(x)=A(x))

(G(F(x))=ln (F(x))-A(x))

(G'(F(x))=frac{1}{F(X)})

(F(x)=F_0(x)-frac{ln(F_0(x)-A(x))}{frac{1}{F_0(x)}}=F_0(x)*(1-ln(F_0(x))+A(x)))

求ln见另一篇blog，qwq

递归求解就行了

因为保证(a_0=0)，而(e^0=1)，所以递归边界返回1就行了

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
using std::vector;
const int mod = 998244353;
const int maxn = 6e5 + 10;
int r[maxn], len;
LL ksm(LL x, LL y) {
	LL re = 1LL;
	while(y) {
		if(y & 1) re = re * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return re;
}
void FNTT(vector<int> &A, int flag) {
	A.resize(len);
	for(int i = 0; i < len; i++) if(i < r[i]) std::swap(A[i], A[r[i]]);
	for(int l = 1; l < len; l <<= 1) {
		int w0 = ksm(3, (mod - 1) / (l << 1));
		for(int i = 0; i < len; i += (l << 1)) {
			int w = 1, a0 = i, a1 = i + l;
			for(int k = 0; k < l; k++, a0++, a1++, w = 1LL * w0 * w % mod) {
				int tmp = 1LL * A[a1] * w % mod;
				A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
				A[a0] = (A[a0] + tmp) % mod;
			}
		}
	}
	if(flag == -1) {
		std::reverse(A.begin() + 1, A.end());
		int inv = ksm(len, mod - 2);
		for(int i = 0; i < len; i++) A[i] = 1LL * inv * A[i] % mod;
	}
}
vector<int> operator * (vector<int> A, vector<int> B) {
	int tot = A.size() + B.size() - 1;
	for(len = 1; len <= tot; len <<= 1);
	for(int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
	FNTT(A, 1), FNTT(B, 1);
	vector<int> ans;
	ans.resize(len);
	for(int i = 0; i < len; i++) ans[i] = 1LL * A[i] * B[i] % mod;
	FNTT(ans, -1);
	ans.resize(tot);
	return ans;
}
vector<int> operator - (const vector<int> &A, const vector<int> &B) {
	vector<int> ans;
	for(int i = 0; i < (int)std::min(A.size(), B.size()); i++) ans.push_back(A[i] - B[i]);
	if(A.size() < B.size()) for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(-B[i]);
	if(A.size() > B.size()) for(int i = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
	return ans;
}
vector<int> operator + (const vector<int> &A, const vector<int> &B) {
	vector<int> ans;
	for(int i = 0; i < (int)std::min(A.size(), B.size()); i++) ans.push_back(A[i] + B[i]);
	if(A.size() < B.size()) for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(B[i]);
	if(A.size() > B.size()) for(int i = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
	return ans;
}
vector<int> inv(const vector<int> &A) {
	if(A.size() == 1) {
		vector<int> ans;
		ans.push_back(ksm(A[0], mod - 2));
		return ans;
	}
	int n = A.size(), _ = (n + 1) >> 1;
	vector<int> B = A, ans;
	ans.push_back(2);
	B.resize(_);
	B = inv(B);
	ans = B * (ans - A * B);
	ans.resize(n);
	return ans;
}
vector<int> getd(const vector<int> &A) {
	vector<int> ans;
	ans.resize(A.size() - 1);
	for(int i = 1; i < (int)A.size(); i++) ans[i - 1] = 1LL * i * A[i] % mod;
	return ans;
}
vector<int> geti(const vector<int> &A) {
	vector<int> ans;
	ans.resize(A.size() + 1);
	for(int i = 1; i < (int)ans.size(); i++) ans[i] = 1LL * A[i - 1] * ksm(i, mod - 2) % mod;
	return ans;
}
vector<int> getln(const vector<int> &A) {
	vector<int> B = inv(A), C = getd(A);
	B = geti(B * C);
	return B;
}
vector<int> gete(const vector<int> &A) {
	if(A.size() == 1) {
		vector<int> ans;
		ans.push_back(1);
		return ans;
	}
	int n = A.size(), _ = (n + 1) >> 1;
	vector<int> ans, B = A;
	ans.push_back(1);
	B.resize(_);
	B = gete(B);
	ans = B * (ans - getln(B) + A);
	ans.resize(n);
	return ans;
}
	
int main() {
	int n = in();
	vector<int> a;
	for(int i = 1; i <= n; i++) a.push_back(in());
	a.resize(a.size() * 2);
	a = gete(a);
	for(int i = 0; i < n; i++) printf("%d%c", a[i], i == n - 1? '
' : ' '); 
	return 0;
}