P5245 【模板】多项式快速幂

思路

调了半天发现ln忘了清空数组了。。。

就是这个式子

\[A^k(x) \equiv e^{k{\ln (A(x)) }}
\]

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 300000;
const int G = 3;
const int invG = 332748118;
const int MOD = 998244353;
int rev[MAXN],inv_val[MAXN];
int pow(int a,int b){
    int ans=1;
    while(b){
        if(b&1)
            ans=(1LL*ans*a)%MOD;
        a=(1LL*a*a)%MOD;
        b>>=1;
    }
    return ans;
}
void cal_rev(int *rev,int n,int lim){
    for(int i=0;i<n;++i)
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(lim-1));
}
void NTT(int *a,int opt,int n,int lim){
    for(int i=0;i<n;++i)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int i=2;i<=n;i<<=1){
        int len=i/2;
        int tmp=pow((opt)?G:invG,(MOD-1)/i);
        for(int j=0;j<n;j+=i){
            int arr=1;
            for(int k=j;k<j+len;k++){
                int t=(1LL*a[k+len]*arr)%MOD;
                a[k+len]=(a[k]-t+MOD)%MOD;
                a[k]=(a[k]+t)%MOD;
                arr=(1LL*arr*tmp)%MOD;
            }
        }
    }
    if(!opt){
        int invN=pow(n,MOD-2);
        for(int i=0;i<n;++i)
            a[i]=(1LL*a[i]*invN)%MOD;
    }
}
void mul(int *a,int *b,int &at,int bt){
    static int tmp1[MAXN];
    int num=(at+bt),n=1,lim=0;
    while(n<=(num+2))
        n<<=1,lim++;
    for(int i=0;i<n;++i)
        tmp1[i]=b[i];
    cal_rev(rev,n,lim);
    NTT(a,1,n,lim);
    NTT(tmp1,1,n,lim);
    for(int i=0;i<n;++i)
        a[i]=(1LL*a[i]*tmp1[i])%MOD;
    NTT(a,0,n,lim);
    at=num;
}
void inv(int *a,int *b,int dep,int &midlen,int &midlim){
    if(dep==1){
        b[0]=pow(a[0],MOD-2);
        return;
    }
    inv(a,b,(dep+1)>>1,midlen,midlim);
    static int tmp[MAXN];
    while((dep<<1)>midlen)
        midlen<<=1,midlim++;
    for(int i=0;i<dep;++i)
        tmp[i]=a[i];
    for(int i=dep;i<midlen;++i)
        tmp[i]=0;
    cal_rev(rev,midlen,midlim);
    NTT(tmp,1,midlen,midlim);
    NTT(b,1,midlen,midlim);
    for(int i=0;i<midlen;++i)
        b[i]=1LL*b[i]*(2-1LL*tmp[i]*b[i]%MOD+MOD)%MOD;
    NTT(b,0,midlen,midlim);
    for(int i=dep;i<midlen;++i)
        b[i]=0;
}
void qd(int *a,int &at){
    for(int i=0;i<at;++i)
        a[i]=(1LL*a[i+1]*(i+1))%MOD;
    a[at]=0;
    at--;
}
void jf(int *a,int &at){
    at++;
    for(int i=at;i>=1;i--)
        a[i]=(1LL*a[i-1]*inv_val[i])%MOD;
    a[0]=0;
}
void ln(int *a,int *b,int &at){
    static int tmp[MAXN];
    int midlen=1,midlim=0,tmpt=at,bt=at;
    for(int i=0;i<=at;++i)
        tmp[i]=a[i];
    inv(a,b,at+1,midlen,midlim);
    qd(tmp,tmpt);
    mul(b,tmp,at,tmpt);
    jf(b,tmpt);
    for(int i=0;i<=bt;i++)
        tmp[i]=0;
    for(int i=bt+1;i<=at;++i)
        tmp[i]=0,b[i]=0;
    at=bt;
}
void exp(int *a,int *b,int dep){
    if(dep==1){
        b[0]=1;
        return;
    }
    exp(a,b,(dep+1)>>1);
    static int tmp1[MAXN];
    for(int i=0;i<dep;++i)
        tmp1[i]=0;
    ln(b,tmp1,dep);
    for(int i=0;i<dep;++i)
        tmp1[i]=(a[i]-tmp1[i]+MOD)%MOD;
    tmp1[0]+=1;
    int midlen=dep-1;
    mul(b,tmp1,midlen,dep-1);
    for(int i=dep;i<midlen;++i)
        b[i]=0;
}
void init_inv(int n){
    inv_val[0]=0;
    inv_val[1]=1;
    for(int i=2;i<=n;i++)
        inv_val[i]=1LL*(MOD-MOD/i)*inv_val[MOD%i]%MOD;
}
void mul(int *a,int n,int k){
    for(int i=0;i<=n;i++)
        a[i]=(1LL*a[i]*k)%MOD;
}
void pow(int *a,int *b,int n,int k){
    static int tmp[MAXN];
    int t=n;
    ln(a,tmp,t);
    mul(tmp,n,k);
    exp(tmp,b,n);
}
int a[MAXN],b[MAXN],n,k;
int main(){
    scanf("%d",&n);
    init_inv(n+10);
    char c=getchar();
    while(c<'0'||c>'9')
        c=getchar();
    while(c>='0'&&c<='9'){
        k=(1LL*k*10%MOD+c-'0')%MOD;
        c=getchar();
    }
    for(int i=0;i<n;++i)
        scanf("%d",&a[i]);
    pow(a,b,n,k);
    for(int i=0;i<n;++i)
        printf("%d ",b[i]);
    return 0;
}