• 多项式全家桶


    namespace Poly{
    const int N=1<<20;
    const int mod=998244353;
    const int G=3;
    int n,rev[N],f[N],g[N],all;
    int ksm(int x,int y){
    int rt=1;
    for(;y;x=(1llxx)%mod,y>>=1)if(y&1)rt=(1llrtx)%mod;
    return rt;
    }
    void NTT(int a,int tp,int LG){
    int lim=(1<<LG),invlim=ksm(lim,mod-2);
    for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(LG-1));
    for(int i=0;i<lim;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    for(int md=1;md<lim;md<<=1){
    int rt=ksm(G,(mod-1)/(md<<1));
    if(tp==-1)rt=ksm(rt,mod-2);
    for(int stp=md<<1,pos=0;pos<lim;pos+=stp){
    int w=1;
    for(int i=0;i<md;i++,w=(1ll
    wrt)%mod){
    int x=a[pos+i],y=(1ll
    wa[pos+md+i])%mod;
    a[pos+i]=(x+y)%mod;
    a[pos+md+i]=(x-y+mod)%mod;
    }
    }
    }
    if(tp==-1)for(int i=0;i<lim;i++)a[i]=(1ll
    a[i]invlim)%mod;
    }
    int A[N],B[N],C[N],D[N],E[N];
    void mul(int a,int b,int c,int LG){//using: Array A and B
    int lim=(1<<LG);
    for(int i=0;i<lim;i++)A[i]=B[i]=0;
    for(int i=0;i<(lim>>1);i++)A[i]=a[i],B[i]=b[i];
    NTT(A,1,LG),NTT(B,1,LG);
    for(int i=0;i<lim;i++)A[i]=1ll
    A[i]
    B[i]%mod;
    NTT(A,-1,LG);
    for(int i=0;i<lim;i++)c[i]=A[i];
    }
    void inv(int a,int b,int LG){//using: Array C
    b[0]=ksm(a[0],mod-2);
    for(int k=1;k<=LG+1;k++){
    mul(b,a,C,k);
    for(int i=0;i<(1<<k);i++)C[i]=(mod-C[i])%mod;
    (C[0]+=2)%=mod;
    mul(C,b,b,k);
    }
    }
    void diff(int a,int b,int lim){
    for(int i=0;i<lim;i++)b[i]=1ll
    a[i+1]
    (i+1)%mod;
    b[lim-1]=0;
    }
    void inte(int a,int b,int lim){
    for(int i=lim-1;i;i--)b[i]=1ll
    a[i-1]
    ksm(i,mod-2)%mod;
    b[0]=0;
    }
    void ln(int a,int b,int LG){//using: Array C
    inv(a,b,LG);
    diff(a,C,1<<LG);
    mul(b,C,b,LG+1);
    inte(b,b,1<<LG);
    }
    void exp(int a,int b,int LG){//using: Array D
    b[0]=1;
    for(int k=1;k<=LG+1;k++){
    ln(b,D,k-1);
    for(int i=0;i<(1<<(k-1));i++)D[i]=(a[i]-D[i]+mod)%mod;
    D[0]=(D[0]+1)%mod;
    mul(b,D,b,k);
    }
    }
    namespace Residue{//Can be commented if the sqrt part has a0=1
    int sqri;
    struct cp{
    int x,y;
    cp(int a=0,int b=0){x=a,y=b;}
    friend cp operator (const cp &a,const cp &b){
    return cp((1ll
    a.x
    b.x%mod+1ll
    a.y
    b.y%mod
    sqri%mod)%mod,(1ll
    a.y
    b.x%mod+1ll
    a.x
    b.y%mod)%mod);
    }
    };
    bool che(int x){
    return ksm(x,mod>>1)==1;
    }
    cp ksm(cp x,int y){
    cp z=cp(1,0);
    for(;y;y>>=1,x=xx)if(y&1)z=zx;
    return z;
    }
    int Cipolla(int x){
    int a=rand()%mod;
    while(true){
    sqri=(1llaa%mod-x+mod)%mod;
    if(!a||che(sqri))a=rand()%mod;
    else break;
    }
    cp tmp=cp(a,1);
    tmp=ksm(tmp,(mod+1)>>1);
    return min(mod-tmp.x,tmp.x);
    }
    }
    using namespace Residue;
    void sqrt(int *a,int *b,int LG){//using:arrays D&E
    b[0]=Cipolla(a[0]);
    for(int k=1;k<=LG+1;k++){
    for(int i=0;i<(1<<(k-1));i++)D[i]=(b[i]<<1)%mod;
    inv(D,E,k-1);
    mul(b,b,b,k);
    for(int i=0;i<(1<<(k-1));i++)(b[i]+=a[i])%=mod;
    mul(b,E,b,k);
    }
    }
    void rever(int *a,int *b,int lim){
    for(int i=0;i<lim;i++)b[i]=a[i];
    reverse(b,b+lim);
    }
    void div(int *a,int *b,int *q,int *r,int lima,int limb){//using:Array D
    rever(b,D,limb);
    for(int i=lima-limb+1;i<limb;i++)D[i]=0;
    int all=0;
    while((1<<all)<lima-limb+1)all++;
    inv(D,q,all);
    rever(a,D,lima);
    all=0;
    while((1<<all)<lima)all++;
    for(int i=lima;i<(1<<all);i++)D[i]=0;
    for(int i=lima-limb+1;i<(1<<all);i++)q[i]=0;
    mul(D,q,q,all+1);
    rever(q,q,lima-limb+1);
    for(int i=lima-limb+1;i<(1<<all);i++)q[i]=0;
    mul(q,b,D,all+1);
    for(int i=0;i<limb-1;i++)r[i]=(f[i]-D[i]+mod)%mod;
    }
    void ksm(int a,int k,int b,int LG){
    ln(a,b,LG);
    for(int i=0;i<(1<<LG);i++)a[i]=1ll
    b[i]
    k%mod;
    exp(a,b,LG);
    }
    }

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  • 原文地址:https://www.cnblogs.com/Troverld/p/14607866.html
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