【Cogs2187】帕秋莉的超级多项式（多项式运算）
题面
题解
多项式运算模板题
只提供代码了。。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define RG register
#define MAX 444444
const int MOD=998244353;
const int Phi=MOD-1;
const int gr=3;
inline int read()
{
RG int x=0,t=1;RG char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
if(ch=='-')t=-1,ch=getchar();
while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
return x*t;
}
int fpow(int a,int b)
{
int s=1;
while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}
return s;
}
int r[MAX],N,l,M;
int Og[MAX];
void NTT(int *P,int opt,int n)
{
for(N=1,l=0;N<n;N<<=1)++l;
for(RG int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
for(RG int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
for(RG int i=1;i<N;i<<=1)
{
RG int W=fpow(gr,Phi/(i<<1));Og[0]=1;
for(RG int j=1;j<i;++j)Og[j]=1ll*Og[j-1]*W%MOD;
for(RG int p=i<<1,j=0;j<N;j+=p)
for(RG int k=0;k<i;++k)
{
RG int X=P[j+k],Y=1ll*Og[k]*P[i+j+k]%MOD;
P[j+k]=(X+Y)%MOD;P[i+j+k]=(X+MOD-Y)%MOD;
}
}
if(opt==-1)
{
reverse(&P[1],&P[N]);
for(RG int i=0,inv=fpow(N,MOD-2);i<N;++i)P[i]=1ll*P[i]*inv%MOD;
}
}
int inv[MAX];
void initinv(int N)
{
inv[0]=inv[1]=1;
for(RG int i=2;i<N;++i)inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
}
int A[MAX],B[MAX];
void Inv(int *a,int *b,int len)
{
if(len==1){b[0]=fpow(a[0],MOD-2);return;}
Inv(a,b,len>>1);
for(RG int i=0;i<len;++i)A[i]=a[i],B[i]=b[i];
NTT(A,1,len<<1);NTT(B,1,len<<1);
for(RG int i=0;i<(len<<1);++i)A[i]=1ll*A[i]*B[i]%MOD*B[i]%MOD;
NTT(A,-1,len<<1);
for(RG int i=0;i<len;++i)b[i]=(b[i]+b[i])%MOD;
for(RG int i=0;i<len;++i)b[i]=(b[i]+MOD-A[i])%MOD;
for(RG int i=0;i<(len<<1);++i)A[i]=B[i]=0;
}
int C[MAX],D[MAX],inv2=fpow(2,MOD-2);
void Sqrt(int *a,int *b,int len)
{
if(len==1){b[0]=sqrt(a[0]);return;}
Sqrt(a,b,len>>1);
for(RG int i=0;i<=len;++i)C[i]=a[i];
Inv(b,D,len);
NTT(C,1,len<<1);NTT(D,1,len<<1);
for(RG int i=0;i<(len<<1);++i)D[i]=1ll*C[i]*D[i]%MOD;
NTT(D,-1,len<<1);
for(RG int i=0;i<len;++i)b[i]=1ll*(b[i]+D[i])%MOD*inv[2]%MOD;
for(RG int i=0;i<=(len<<1);++i)C[i]=D[i]=0;
}
void Dao(int *a,int *b,int len)
{
for(RG int i=1;i<len;++i)b[i-1]=1ll*i*a[i]%MOD;
b[len]=b[len-1]=0;
}
void Jifen(int *a,int *b,int len)
{
for(RG int i=1;i<len;++i)b[i]=1ll*a[i-1]*inv[i]%MOD;
b[0]=0;
}
void Getln(int *a,int *b,int len)
{
int A[MAX],B[MAX];
memset(A,0,sizeof(A));memset(B,0,sizeof(B));
Dao(a,A,len);
Inv(a,B,len);
NTT(A,1,len<<1);NTT(B,1,len<<1);
for(RG int i=0;i<(len<<1);++i)A[i]=1ll*A[i]*B[i]%MOD;
NTT(A,-1,len<<1);
Jifen(A,b,len);
for(RG int i=0;i<(len<<1);++i)A[i]=B[i]=0;
}
int E[MAX];
void Exp(int *a,int *b,int len)
{
if(len==1){b[0]=1;return;}
Exp(a,b,len>>1);
for(RG int i=0;i<len;++i)D[i]=b[i];
Getln(b,E,len);
for(RG int i=0;i<len;++i)E[i]=(MOD-E[i]+a[i])%MOD;E[0]=(E[0]+1)%MOD;
NTT(E,1,len<<1);NTT(D,1,len<<1);
for(RG int i=0;i<(len<<1);++i)D[i]=1ll*D[i]*E[i]%MOD;
NTT(D,-1,len<<1);
for(RG int i=0;i<len;++i)b[i]=D[i];
for(RG int i=0;i<(len<<1);++i)D[i]=E[i]=0;
}
void FastPow(int *a,int *b,int K,int len)
{
int E[MAX];memset(E,0,sizeof(E));
Getln(a,E,len);
for(RG int i=0;i<len;++i)E[i]=1ll*K*E[i]%MOD;
Exp(E,b,len);
}
int X[MAX],Y[MAX];
int n,K;
int main()
{
freopen("polynomial.in","r",stdin);
freopen("polynomial.out","w",stdout);
n=read();K=read();
for(RG int i=0;i<n;++i)X[i]=read();
int N;for(N=1;N<=n;N<<=1);initinv(N);
Sqrt(X,Y,N);memset(X,0,sizeof(X));//for(int i=0;i<n;++i)printf("%d ",Y[i]);puts("");
Inv(Y,X,N);memset(Y,0,sizeof(Y));//for(int i=0;i<n;++i)printf("%d ",X[i]);puts("");
Jifen(X,Y,N);memset(X,0,sizeof(X));//for(int i=0;i<n;++i)printf("%d ",Y[i]);puts("");
Exp(Y,X,N);memset(Y,0,sizeof(Y));//for(int i=0;i<n;++i)printf("%d ",X[i]);puts("");
Inv(X,Y,N);memset(X,0,sizeof(X));Y[0]=(Y[0]+1)%MOD;//for(int i=0;i<n;++i)printf("%d ",Y[i]);puts("");
Getln(Y,X,N);memset(Y,0,sizeof(Y));X[0]=(X[0]+1)%MOD;//for(int i=0;i<n;++i)printf("%d ",X[i]);puts("");
FastPow(X,Y,K,N);memset(X,0,sizeof(X));
Dao(Y,X,n);for(RG int i=0;i<n;++i)printf("%d ",X[i]);puts("");
return 0;
}