luogu4726.cpp

// 多项式exp, Newton's method
// 调不出来..

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1e5+5;
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch = getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}

const int P = 998244353;

inline int qpow(int x,int y){int res(1);while(y){if(y&1) res = 1ll*res*x%P;x = 1ll*x*x%P;y>>=1;}return res;}

int rev[MAXN<<2];

void change(int y[],int len){
    for(int i=0;i<len;++i){
        rev[i] = rev[i>>1]>>1;
        if(i&1) rev[i] |= len>>1;
    }
    for(int i=0;i<len;++i)
        if(i<rev[i]) swap(y[i],y[rev[i]]);
}

void ntt(int y[],int len,int on){
    change(y,len);
    for(int h=2;h<=len;h<<=1){
        int gn = qpow(3,(P-1)/h);
        for(int j=0;j<len;j+=h){
            int g = 1;
            for(int k=j;k<j+h/2;++k){
                int u = y[k];
                int t = 1ll*g*y[k+h/2]%P;
                y[k] = (u+t)%P;
                y[k+h/2] = (u-t+P)%P;
                g = 1ll*g*gn%P;
            }
        }
    }
    if(on==-1) {
        reverse(y+1,y+len);
        register int inv = qpow(len,P-2);
        for(int i=0;i<len;++i) y[i] = 1ll*y[i]*inv%P;
    }
}

int lnf[MAXN<<2];

static int inv_t[MAXN<<2];

void get_inv(int f[],int g[],int n){
    // 递归:把两个n/2的合并成一个n的
    if(n==1) {f[0] = qpow(g[0],P-2);return;}
    get_inv(f,g,(n+1)>>1);
    int len(1);
    while(len<(n<<1)) len<<=1;
    copy(g,g+n,inv_t);
    fill(inv_t+n,inv_t+len,0); // 自带清空

    ntt(f,len,1);
    ntt(inv_t,len,1);
    for(int i=0;i<len;++i){
        f[i] = 1ll*f[i]*(2-1ll*f[i]*inv_t[i]%P+P)%P;
    }
    ntt(f,len,-1);
    fill(f+n,f+len,0);  // 取模
}

static int temp2[MAXN<<2];

// 求对数
void ln(int f[],int g[],int n){
    get_inv(f,g,n);
    for(int i=1;i<n;++i)
        temp2[i-1]=1ll*g[i]*i%P;
    int len(1);
    while(len<(n<<1)) len<<=1;
    fill(temp2+n-1,temp2+len,0);
    ntt(temp2,len,1),ntt(f,len,1);
    for(int i=0;i<len;++i)
        f[i] = 1ll*f[i]*temp2[i]%P;
    ntt(f,len,-1);
    for(int i=1;i<n;++i)
        f[i]=1ll*qpow(i,P-2)*f[i-1]%P;
}

static int temp[MAXN<<2];

void solve(int f[],int g[],int n){
    // 递归:把两个n/2的合并成一个n的
    if(n==1) {f[0] = 1;return;}
    solve(f,g,(n+1)>>1);
    int len(1);
    while(len<(n<<1)) len<<=1;
    copy(g,g+n,temp);
    fill(temp+n,temp+len,0);
    ln(lnf,f,len); // 这个大小是模n意义下的..
    lnf[0] = (temp[0]-lnf[0]+1+P)%P;
    for(int i=1;i<len;++i){
        lnf[i] = (temp[i]-lnf[i]+P)%P;
    }
    ntt(f,len,1);
    ntt(lnf,len,1);
    for(int i=0;i<len;++i)
        f[i] = 1ll*f[i]*lnf[i]%P;
    ntt(f,len,-1);
    fill(f+n,f+len,0);  // 取模
}

int f[MAXN<<2],g[MAXN<<2];

int main(){
    int n=read();
    for(int i=0;i<n;++i) g[i]=read();
    solve(f,g,n);
    for(int i=0;i<n;++i)
        printf("%d%c",f[i],i==n-1?'\n':' ');
}