洛谷P5264 【模板】多项式三角函数

题面

传送门

题解

据说有一个叫做欧拉公式的东西

[e^{ix}=cos(x)+isin(x) ]

别问我为啥我今天第一次看到它

那么显然也有

[e^{-ix}=cos(x)-isin(x) ]

两个柿子相加得到

[e^{ix}+e^{-ix}=2cos(x) ]

[cos(x)={e^{ix}+e^{-ix}over 2} ]

同理可得

[sin(x)={e^{ix}-e^{-ix}over 2i} ]

然后左转抄板……

等会儿这里这个(i)咋整？

因为(i^2=-1)，那么在模(998244353)意义下有

[i^2equiv -1equiv 998244352mod(998244353) ]

爆搜之后可以解得(iequiv 86583718)

然后没有然后了

//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]=' ';
}
const int N=(1<<18)+5,P=998244353,img=86583718,inv2=499122177,iimg=954952494;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
    R int res=1;
    for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
    return res;
}
int inv[N],r[21][N],rt[2][N<<1],lg[N],lim,d;
int iinv(R int x){return x<=262144?inv[x]:mul(P-P/x,iinv(P%x));}
void Pre(){
    fp(d,1,18){
        fp(i,1,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1));
        lg[1<<d]=d;
    }
    inv[0]=inv[1]=1;
    fp(i,2,262144)inv[i]=mul(P-P/i,inv[P%i]);
    for(R int t=(P-1)>>1,i=1,x,y;i<=262144;i<<=1,t>>=1){
        x=ksm(3,t),y=iinv(x),rt[0][i]=rt[1][i]=1;
        fp(k,1,i-1)
            rt[1][i+k]=mul(rt[1][i+k-1],x),
            rt[0][i+k]=mul(rt[0][i+k-1],y);
    }
}
void NTT(int *A,int ty){
    fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]);
    for(R int mid=1;mid<lim;mid<<=1)
        for(R int j=0,t;j<lim;j+=(mid<<1))
            fp(k,0,mid-1)
                A[j+k+mid]=dec(A[j+k],t=mul(rt[ty][mid+k],A[j+k+mid])),
                A[j+k]=add(A[j+k],t);
    if(!ty)fp(i,0,lim-1)A[i]=mul(A[i],inv[lim]);
}
void Inv(int *a,int *b,int len){
    if(len==1)return b[0]=iinv(a[0]),void();
    Inv(a,b,len>>1);
    static int A[N],B[N];lim=(len<<1),d=lg[lim];
    fp(i,0,len-1)A[i]=a[i],B[i]=b[i];
    fp(i,len,lim-1)A[i]=B[i]=0;
    NTT(A,1),NTT(B,1);
    fp(i,0,lim-1)A[i]=mul(A[i],mul(B[i],B[i]));
    NTT(A,0);
    fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),A[i]);
    fp(i,len,lim-1)b[i]=0;
}
void Ln(int *a,int *b,int len){
    static int A[N],B[N];
    fp(i,1,len-1)A[i-1]=mul(a[i],i);A[len-1]=0;
    Inv(a,B,len);lim=(len<<1),d=lg[lim];
    fp(i,len,lim-1)A[i]=B[i]=0;
    NTT(A,1),NTT(B,1);
    fp(i,0,lim-1)A[i]=mul(A[i],B[i]);
    NTT(A,0);
    fp(i,1,len-1)b[i]=mul(A[i-1],inv[i]);b[0]=0;
    fp(i,len,lim-1)b[i]=0;
}
void Exp(int *a,int *b,int len){
    if(len==1)return b[0]=1,void();
    Exp(a,b,len>>1);
    static int A[N];Ln(b,A,len);
    lim=(len<<1),d=lg[lim];
    A[0]=dec(a[0]+1,A[0]);
    fp(i,1,len-1)A[i]=dec(a[i],A[i]);
    fp(i,len,lim-1)A[i]=b[i]=0;
    NTT(A,1),NTT(b,1);
    fp(i,0,lim-1)b[i]=mul(A[i],b[i]);
    NTT(b,0);
    fp(i,len,lim-1)b[i]=0;
}
int A[N],B[N],n,ty;
void cos(int *a,int *b,int len){
    static int p[N],A[N],B[N];
    fp(i,0,len-1)p[i]=mul(img,a[i]);
    Exp(p,A,len);
    Inv(A,B,len);
    fp(i,0,len-1)b[i]=mul(add(A[i],B[i]),inv2);
}
void sin(int *a,int *b,int len){
    static int p[N],A[N],B[N];
    fp(i,0,len-1)p[i]=mul(img,a[i]);
    Exp(p,A,len);
    Inv(A,B,len);
    fp(i,0,len-1)b[i]=mul(dec(A[i],B[i]),iimg);
}
int main(){
//	freopen("testdata.in","r",stdin);
    n=read(),ty=read(),Pre();
    fp(i,0,n-1)A[i]=read();
    int len=1;while(len<=n)len<<=1;
    if(ty)cos(A,B,len);else sin(A,B,len);
    fp(i,0,n-1)print(B[i]);
    return Ot(),0;
}