【模板】分治FFT

题意

分析

如果我们已经求得了 f[L],f[L+1] ... f[mid],他们均能对f[mid+1],f[mid+2]...,f[R]产生贡献

对于x ∈ [mid+1,r] f[x] += sum_{i=L}^{mid}(f[i]*g[x-i])

等式右边满足卷积模式

具体看代码

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
ll gg[400005],f[400005],A[400005],B[400005];
int g=3,mod=998244353;
ll qpow(ll a,ll b) {
    ll res=1;
    while(b) {
        if(b&1) res = res * a % mod;
        a = a*a %mod;
        b >>= 1;
    }
    return res;
}
int rev(int x,int r) {
    int ans=0;
    for(int i=0;i<r;i++) if(x&(1<<i)) ans+=1<<(r-i-1);
    return ans;
}
void ntt(int n,ll a[],int on) {
    int r = 0;
    while((1<<r)<n) r++;
    for(int i=0;i<n;i++) {
        int tmp = rev(i,r);
        if(i < tmp)  swap(a[i],a[tmp]);
    }
    for(int s=1;s<=r;s++) {
        int m = 1<<s;
        ll wn=qpow(g,(mod-1)/m);
        for(int k=0;k<n;k+=m){
            ll w=1;
            for(int j=0;j<(m>>1);j++) {
                ll t = w * a[k+j+(m>>1)] % mod;
                ll u=a[k+j];
                a[k+j] = (u+t)%mod;
                a[k+j+(m>>1)] = (u-t)%mod;if(a[k+j+(m>>1)]<0)a[k+j+(m>>1)]+=mod;
                w = w*wn%mod;
            }
        }
    }
    if(on==-1) {
        for(int i=1;i<(n>>1);i++) swap(a[i],a[n-i]);
        ll inv = qpow(n,mod-2);
        for(int i=0;i<n;i++) a[i] = a[i] * inv % mod;
    }
}
void solve(int l,int r) {
   if(l == r) return;
   int mid = (r+l)>>1;
   solve(l,mid);
   int siz = r-l+1;
   int len = 1;
   while(len < siz+siz) len <<=1;
   for(int i=0;i<len;i++) A[i] = B[i] = 0;
   for(int i=l;i<=mid;i++) A[i-l] = f[i];
   for(int i=1;i<=r-l;i++) B[i] = gg[i];
    ntt(len,A,1);ntt(len,B,1);
    for(int i=0;i<len;i++) A[i]=A[i]*B[i]%mod;
    ntt(len,A,-1);
    for(int i=mid+1;i<=r;i++) f[i] = (f[i] + A[i-l])%mod;
    solve(mid+1,r);
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i=1;i<n;i++) {
        cin >> gg[i];
        gg[i] %= mod;
    }
    f[0] = 1;
    solve(0,n-1);
    for(int i=0;i<n;i++) cout<<f[i]<<" ";
    cout<<"
";
}