多项式取 ln

模板题传送门

给你 \(A\)，让你求 \(B(x) \equiv lnA(x) \pmod {x^n}\)

两遍求导，得到 \(B'(x) \equiv \frac{A'(x)}{A(x)} \pmod {x^n}\)

然后积分即有 \(B(x) \equiv \int\frac{A'(x)}{A(x)}dx \pmod {x^n}\)

#include
using namespace std;

#define debug(x) cerr << #x << ": " << (x) << endl
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)

using pii = pair;
using ll = long long;

#define int long long

inline void read(int &x){
    int s=0; x=1;
    char ch=getchar();
    while(ch<'0' || ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0' && ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=3e5+5, rt=3, mod=998244353;

int rev[N], tot=1, bit;

ll fpow(ll x, int p, ll mod){
	int res=1;
	for(; p; p>>=1, x=x*x%mod) if(p&1) res=res*x%mod;
	return res;
}

ll inv(ll x, ll mod){
	return fpow(x, mod-2, mod);
}

ll mul(ll x, int p, ll mod){
	ll res=0;
	for(; p; p>>=1, x=(x+x)%mod) if(p&1) res=(res+x)%mod;
	return res;
}

void NTT(ll *a, int type, int mod){
	for(int i=0; i>1, a, b);
	
	// init
	bit=0, tot=1;
	while(tot<=(sz-1<<1)) tot<<=1, bit++;
	for(int i=0; i>1]>>1)|((i&1)<<(bit-1));
	
	rep(i,0,sz-1) C[i]=a[i];
	rep(i,sz,tot-1) C[i]=0;
	
	NTT(C, 1, mod), NTT(b, 1, mod);
	rep(i,0,tot-1) b[i]=(2-C[i]*b[i]%mod+mod)%mod*b[i]%mod;
	NTT(b, -1, mod);
	
	rep(i,sz,tot-1) b[i]=0;
}

void poly_der(int sz, int *a, int *b){
	rep(i,1,sz-1) b[i-1]=i*a[i]%mod;
	b[sz-1]=0;
}

void poly_int(int sz, int *a, int *b){
	rep(i,1,sz-1) b[i]=a[i-1]*inv(i, mod)%mod;
	b[0]=0; 
}

int inv_A[N], der_A[N];

void conv(int *a, int *b){
	NTT(a, 1, mod), NTT(b, 1, mod);
	rep(i,0,tot-1) a[i]=a[i]*b[i]%mod;
	NTT(a, -1, mod);
}

void poly_ln(int sz, int *a, int *b){
	poly_inv(sz, a, inv_A);
	poly_der(sz, a, der_A);
	
	// init
	bit=0, tot=1;
	while(tot<=sz-1+sz-2) tot<<=1, bit++;
	for(int i=0; i>1]>>1)|((i&1)<<(bit-1));

	conv(inv_A, der_A);
	
	poly_int(sz, inv_A, b);
}

signed main(){
	cin>>n;
	rep(i,0,n-1) read(A[i]);
	poly_ln(n, A, B);
	rep(i,0,n-1) cout<