多项式相关模板#1

一个模板,功能大概就是加减,乘法,求逆和除法,后面(可能)会加上exp,ln,多点求值,插值,取模

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 280010
#define M 998244353
#define LL long long
using namespace std;
int W[N],iW[N];
inline LL pow(LL x,LL k,LL s=1){
    for(;k;x=x*x%M,k>>=1) k&1?s=s*x%M:0;
    return s;
}
inline void init(){
    for(int j=1;j<=N;j<<=1){
        W[j]=pow(3,(M-1)/j);
        iW[j]=pow(W[j],M-2);
    }
}
inline void NTT(int* A,int n,int g){
    static int b[N];
    for(int i=0,j,k,t;i<n;++i){
        for(j=0,k=i,t=n-1;t;t>>=1,k>>=1) 
            j=(j<<1)|(k&1);
        b[j]=A[i];
    }
    for(int m=2;m<=n;m<<=1){
        LL w=g?W[m]:iW[m],u,v,z;
        for(int i,k=m>>1,j=0;j<n;j+=m)
            for(z=1,i=0;i<k;++i,z=z*w%M){
                u=b[i+j]; v=z*b[i+j+k]%M;
                b[i+j]=(u+v)%M; b[i+j+k]=(u-v+M)%M;
            }
    }
    if(!g){
        g=pow(n,M-2);
        for(int i=0;i<n;++i) A[i]=(LL)b[i]*g%M;
    } else memcpy(A,b,n<<2);
}
struct poly{
    int n,a[N];
    inline int len(){ return n; }
    inline int& operator[] (int i){ return a[i]; }
    inline void trans(int* b,int m){
        n=m; memcpy(a,b,m+1<<2);
    }
};
inline poly operator+ (poly a,poly b){
    poly c; c.n=max(a.n,b.n);
    for(int i=0;i<=c.n;++i)
        c[i]=(a[i]+b[i])%M;
    return c;
}
inline poly operator- (poly a,poly b){
    poly c; c.n=max(a.n,b.n);
    for(int i=0;i<=c.n;++i)
        c[i]=(a[i]-b[i]+M)%M;
    return c;
}
inline poly operator* (poly a,poly b){
    poly c; c.n=a.n+b.n;
    if((LL)a.n*b.n<=-1){
        for(int i=0;i<=a.n;++i)
            for(int j=0;j<=b.n;++j)
                c[i+j]=(c[i+j]+(LL)a[i]*b[j])%M;
        return c;
    }
    int n=1; for(;n<=c.n;n<<=1);
    static int A[N],B[N];
    memset(A,0,n<<2);
    memset(B,0,n<<2);
    memcpy(A,a.a,a.n+1<<2);
    memcpy(B,b.a,b.n+1<<2);
    NTT(A,n,1); NTT(B,n,1);
    for(int i=0;i<n;++i) A[i]=(LL)A[i]*B[i]%M;
    NTT(A,n,0);
    c.trans(A,c.n); return c;
}
int v[N],iA[N],iB[N];
inline void gInv(int n){
    if(n==1){ *iA=pow(*v,M-2); return; }
    gInv(n+1>>1);
    int m=1; for(;m<n+n+3;m<<=1);
    for(int i=n;i<m;++i) iA[i]=iB[i]=0;
    memcpy(iB,v,n<<2);
    NTT(iA,m,1); NTT(iB,m,1);
    for(int i=0;i<m;++i)
        iA[i]=iA[i]*(M+2-((LL)iA[i]*iB[i]%M))%M;
    NTT(iA,m,0);
    for(int i=n;i<m;++i) iA[i]=0;
}
inline poly operator~ (poly x){
    poly y; y.n=x.n;
    for(int i=0;i<=x.n;++i) v[i]=x[i];
    gInv(x.n+1);
    y.trans(iA,x.n);
    return y;
}
inline void div(poly a,poly b,poly& d,poly& r){
    int n=a.n,m=b.n;
    if(n<m){ d.n=d[0]=0; r=a; return; }
    int len=1; for(;len<n+n;len<<=1);
    poly ra=a,rb=b;
    reverse(ra.a,ra.a+n+1);
    reverse(rb.a,rb.a+m+1);
    for(int i=n-m+1;i<=rb.n;++i) rb[i]=0;
    rb.n=n-m; rb=~rb;
    d=ra*rb; d.n=n-m; reverse(d.a,d.a+d.n+1);
    r=a-(b*d); for(r.n=m;!r[r.n];r.n--);
}
inline poly operator/ (poly a,poly b){poly d,r; div(a,b,d,r); return d;}
inline poly operator% (poly a,poly b){poly d,r; div(a,b,d,r); return r;}

int main(){ init(); } 

posted @ 2018-04-23 16:30  扩展的灰(Extended_Ash)  阅读(178)  评论(0编辑  收藏  举报