• 多项式板子


    咱的多项式板子,从(NTT)到反三角函数应有尽有

    常数应该比较小,你谷上的板子提交里一般能跑到第一页的

    似乎我就没见过几个人现在还在用数组写多项式的了……

    //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;
    }
    void Sqrt(int *a,int *b,int len){
        if(len==1)return b[0]=1,void();
        Sqrt(a,b,len>>1);
        static int A[N],B[N];
        fp(i,0,len-1)A[i]=a[i];Inv(b,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,0,len-1)b[i]=mul(add(b[i],A[i]),inv2);
        fp(i,len,lim-1)b[i]=0;
    }
    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);
    }
    void Arcsin(int *a,int *b,int len){
        static int A[N],B[N];
        lim=(len<<1),d=lg[lim];
        fp(i,0,len-1)A[i]=a[i];fp(i,len,lim-1)A[i]=0;
        NTT(A,1);fp(i,0,lim-1)A[i]=mul(A[i],A[i]);
        NTT(A,0);fp(i,0,len-1)A[i]=P-A[i];++A[0];
        Sqrt(A,B,len),Inv(B,A,len);
        lim=(len<<1),d=lg[lim];
        fp(i,1,len-1)B[i-1]=mul(a[i],i);B[len-1]=0;
        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;
    }
    void Arctan(int *a,int *b,int len){
        static int A[N],B[N];
        lim=(len<<1),d=lg[lim];
        fp(i,0,len-1)A[i]=a[i];fp(i,len,lim-1)A[i]=0;
        NTT(A,1);fp(i,0,lim-1)A[i]=mul(A[i],A[i]);
        NTT(A,0);++A[0];
        Inv(A,B,len);
        lim=(len<<1),d=lg[lim];
        fp(i,1,len-1)A[i-1]=mul(a[i],i);A[len-1]=0;
        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;
    }
    int main(){Pre();}
    
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  • 原文地址:https://www.cnblogs.com/bztMinamoto/p/10579318.html
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