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  • 分治 FFT 模板的三种过法

    生成函数做法

    [egin{align} f[i] &= sum_{j=1}^i f[i-j]g[j] \ f[i] & =sum_{j=0}^{i-1}f[(i-1)-j]g[j+1] end{align} ]

    把 g 整体左移一位后:

    [egin{align} f[i] & =sum_{j=0}^{i-1}f[(i-1)-j]g[j] \ f[i] &= (f*g)[i-1] \ &i > 0 end{align} ]

    (f(z) = z*f(z)*g(z) + 1)

    这相当于 g 不需要左移。

    于是 (f(z) = f(z)g(z) + 1)

    在膜 (x^n) 意义下, (f(z) = dfrac 1{1-g(z)})

    #include<bits/stdc++.h>
    typedef long long LL;
    using namespace std;
    const int N = 4e5 + 233, mo = 998244353;
    LL ksm(LL a, LL b) {
    	LL res = 1ll;
    	for(; b; b>>=1, a=a*a%mo)
    		if(b & 1) res = res*a%mo;
    	return res;
    }
    const int g = 3, ig = ksm(g, mo - 2);
    
    int rv[N];
    void NTT(LL *a, int n, int type) {
    	for(int i = 0; i < n; ++i) if(i < rv[i]) swap(a[i], a[rv[i]]);
    	for(int m = 2; m <= n; m = m << 1) {
    		LL w = ksm(type == 1 ? g : ig, (mo - 1) / m);
    		for(int i = 0; i < n; i += m) {
    			LL tmp = 1ll;
    			for(int j = 0; j < (m >> 1); ++j) {
    				LL p = a[i + j], q = tmp * a[i + j + (m >> 1)] % mo;
    				a[i + j] = (p + q) % mo, a[i + j + (m >> 1)] = (p - q + mo) % mo;
    				tmp = tmp * w % mo;
    			}
    		}
    	}
    	if(type == -1) {
    		LL Inv = ksm(n, mo - 2);
    		for(int i = 0; i < n; ++i) a[i] = a[i] * Inv % mo;
    	}
    }
    
    int n;
    LL G[N], f[N], t[N];
    
    void poly_inv(int deg, LL *a, LL *b) {
    	if(deg == 1) { b[0] = ksm(a[0], mo - 2); return;}
    	poly_inv((deg + 1) >> 1, a, b);
    	int len = 1; while(len < (deg << 1)) len = len << 1;
    	for(int i = 0; i < deg; ++i) t[i] = a[i];
    	for(int i = deg; i < len; ++i) t[i] = 0ll;
    	for(int i = 1; i < len; ++i) rv[i] = (rv[i>>1]>>1) | (i&1 ? len>>1 : 0);
    	NTT(t, len, 1), NTT(b, len, 1);
    	for(int i = 0; i < len; ++i) b[i] = b[i] * (2ll - t[i] * b[i] % mo) % mo;
    	NTT(b, len, -1);
    	for(int i = deg; i < len; ++i) b[i] = 0ll;
    }
    
    int main() {
    	scanf("%d", &n);
    	for(int i = 1; i <= n-1; ++i) scanf("%lld", &G[i]), G[i]%=mo, G[i] = (mo - G[i]);
    	G[0] = 1ll;
    	poly_inv(n, G, f);
    	for(int i = 0; i < n; ++i) cout << (f[i] % mo + mo) % mo << ' ';
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/tztqwq/p/14334166.html
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