P4721_分治NTT

NTT那块还要再看一下。。。

#include <cstdio>
#include <iostream>
#include <cmath>

typedef long long ll;
const int NR = 1 << 22, g = 3, gi = 332748118, mod = 998244353;

using namespace std;
int n, m, rev[NR];
ll a[NR], b[NR],c[NR],d[NR];
ll quikpow(ll a, ll k)
{
    ll res = 1;
    while (k > 0)
    {
        if (k&1)res=res*a%mod;
        a=a*a%mod;
        k>>=1;
    }
    return res;
}

void NTT(ll *a,int n,int type){
    for(int i=0;i<=n;++i){
        if(i<rev[i])swap(a[i],a[rev[i]]);
    }
    for(int i=1;i<n;i<<=1){
        ll gn=quikpow(type?g:gi,(mod-1)/(i<<1));
        for(int j=0;j<n;j+=(i<<1)){
            ll g0=1;
            for(int k=0;k<i;++k,g0=g0*gn%mod){
                ll x=a[j+k],y=g0*a[i+j+k]%mod;
                a[j+k]=(x+y)%mod;
                a[i+j+k]=(x-y+mod)%mod;
            }
        }
    }
    if(type==1)return ;
    ll inv =  quikpow(n, mod - 2);
    //reverse(a+1,a+n);
    for(int i=0;i<n;++i)a[i]=a[i]*inv%mod;
}

void cdq(int l,int r){
    if(l==r)return;
    int mid=(l+r)>>1;
    cdq(l,mid);

    int ttt=(r-l+1)<<1,len=1<<max((int)ceil(log2(ttt)),1);
    for(int i=0;i<len;++i){
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(max((int)ceil(log2(ttt)),1)-1));
    }

    for(int i=0;i<len;++i)c[i]=d[i]=0;
    for(int i=l;i<=r;++i)c[i-l]=a[i-l];
    for(int i=l;i<=mid;++i)d[i-l]=b[i];
    NTT(c,len,1);
    NTT(d,len,1);
    for(int i=0;i<len;++i)c[i]=c[i]*d[i]%mod;
    NTT(c,len,0);
    for(int i=mid+1;i<=r;++i)b[i]=(b[i]+c[i-l])%mod;
    cdq(mid+1,r);
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<n;++i)scanf("%lld",a+i);
    b[0]=1;
    cdq(0,n-1);
    
    for(int i=0;i<n;++i)printf("%lld ",b[i]);
    return 0;
}