P4721 【模板】分治 FFT

(color{#0066ff}{ 题目描述 })

给定长度为 (n-1) 的数组 (g[1],g[2],..,g[n-1])，求 (f[0],f[1],..,f[n-1])，其中

(f[i]=sum_{j=1}^if[i-j]g[j])

边界为 (f[0]=1) 。答案模 (998244353) 。

(color{#0066ff}{输入格式})

第一行一个正整数 (n) 。

第二行共 (n-1) 个非负整数 (g[1],g[2],..,g[n-1])，用空格隔开。

(color{#0066ff}{输出格式})

一行共 (n) 个非负整数，表示 (f[0],f[1],..,f[n-1]) 模 (998244353) 的值。

(color{#0066ff}{输入样例})

4
3 1 2
    
10
2 456 32 13524543 998244352 0 1231 634544 51

(color{#0066ff}{输出样例})

1 3 10 35
    
1 2 460 1864 13738095 55389979 617768468 234028967 673827961 708520894

(color{#0066ff}{数据范围与提示})

(2≤n≤10^5)

(0leq g[i]<998244353)

(color{#0066ff}{题解})

然而这题可以用多项式求逆过(雾

显然可以看出(f*g=f-f_0)

然后。。。(f=frac{1}{1-g})

求个逆就没了。。

#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;
}
const int maxn = 4e5 + 10;
const int mod = 998244353;
using std::vector;
int n, 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 FFT(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 * w * A[a1] % mod;
				A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
				A[a0] = (A[a0] + tmp) % mod;
			}
		}
	}
	if(!(~flag)) {
		std::reverse(A.begin() + 1, A.end());
		int inv = ksm(len, mod - 2);
		for(int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % 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));
	FFT(A, 1), FFT(B, 1);
	vector<int> ans;
	for(int i = 0; i < len; i++) ans.push_back(1LL * A[i] * B[i] % mod);
	FFT(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]);
	for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(-B[i]);
	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;
	}
	vector<int> ans, B = A;
	int n = A.size(), _ = (n + 1) >> 1;
	B.resize(_);
	ans.push_back(2);
	B = inv(B);
	ans = B * (ans - A * B);
	ans.resize(n);
	return ans;
}
int main() {
	int n = in();
	vector<int> a;
	a.push_back(1);
	for(int i = 1; i < n; i++) a.push_back(mod - in());
	a = inv(a);
	for(int i = 0; i < n; i++) printf("%d%c", a[i], i == n - 1? '
' : ' ');
	return 0;
}