\(\color{#0066ff}{ 题目描述 }\)
给出 \(n-1\) 次多项式 \(A(x)\)，求一个 \(\bmod{\:x^n}\)下的多项式 \(B(x)\)，满足 \(B(x) \equiv \ln A(x)\)
在 \(\text{mod } 998244353\)下进行，且 \(a_i \in [0, 998244353] \cap \mathbb{Z}\)
\(\color{#0066ff}{输入格式}\)
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 \(a_0 = 1\).
\(\color{#0066ff}{输出格式}\)
输出 \(n\) 个整数，表示答案多项式中的系数 \(a_0, a_1, \cdots, a_{n-1}\).
\(\color{#0066ff}{输入样例}\)
6
1 927384623 878326372 3882 273455637 998233543\(\color{#0066ff}{输出样例}\)
0 927384623 817976920 427326948 149643566 610586717\(\color{#0066ff}{数据范围与提示}\)
对于 \(100\%\)的数据，\(n \le 10^5\).
\(\color{#0066ff}{ 题解 }\)
推一下式子
\(B(x)=\ln(A(x))\)
\(B'(x)=\frac{A'(x)}{A(x)}\)
\(B(x)=\int \frac{A'(x)}{A(x)}\)
积分相当于逆求导，由于是多项式，很好求
先求个逆，再求个导，再FFT一下，再积分一下就行了
#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 = 4e5 + 10;
int len, r[maxn];
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 * w * w0 % 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 = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
if(A.size() < B.size()) for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(-B[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> ans, B = A;
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;
}
int main() {
int n = in();
vector<int> a, b, c;
for(int i = 0; i < n; i++) a.push_back(in());
b = inv(a);
a = getd(a);
c = a * b;
c = geti(c);
for(int i = 0; i < n; i++) printf("%d%c", c[i], i == n - 1? '\n' : ' ');
return 0;
}