P4245 【模板】任意模数NTT

Luogu4245

只要做三次的NTT,快的飞起

普通NTT,做9次

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
    register LL x=0,f=1;register char c=getchar();
    while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
    while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
    return f*x;
}

namespace Math{
    inline int qpow(LL a,int b,int mod){
        LL res=1;
        while(b){
            if(b&1) (res*=a)%=mod;
            (a*=a)%=mod;
            b>>=1;
        }
        return (int)res;
    }
    inline int inv(int x,int mod){return qpow(x,mod-2,mod);}
}using namespace Math;

const int mod1=998244353,mod2=1004535809,mod3=469762049,G=3;
const LL mod_1_2=(LL)mod1*mod2;
const int inv_1=inv(mod1,mod2),inv_1_2=inv(mod_1_2%mod3,mod3);
const int MAXN=3e5+5;

int mod;

struct Int{
    int A,B,C;
    inline Int(){}
    inline Int(int _num):A(_num),B(_num),C(_num){}
    inline Int(int _A,int _B,int _C):A(_A),B(_B),C(_C){}
    static inline Int add(Int x){///static不可省
        return Int(x.A+(x.A>>31&mod1),x.B+(x.B>>31&mod2),x.C+(x.C>>31&mod3));
    }
    inline friend Int operator + (Int x,Int y){
        return add(Int(x.A+y.A-mod1,x.B+y.B-mod2,x.C+y.C-mod3));
    }
    inline friend Int operator - (Int x,Int y){
        return add(Int(x.A-y.A,x.B-y.B,x.C-y.C));
    }
    inline friend Int operator * (Int x,Int y){
        return Int((LL)x.A*y.A%mod1,(LL)x.B*y.B%mod2,(LL)x.C*y.C%mod3);
    }
    inline int get(){
        LL x=(B-A+mod2)%mod2*inv_1%mod2*mod1+A;
        return ((C-x%mod3+mod3)%mod3*inv_1_2%mod3*(mod_1_2%mod)%mod+x)%mod;
    }
}A[MAXN],B[MAXN];

namespace N_T_T{
    Int Wn[MAXN],ilim;
    int rev[MAXN],limit=1,l;
    inline void init(int n){
        limit=1,l=0;
        while(limit<=n) limit<<=1,l++;
        for(int i=0;i<limit;i++)
            rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
        Int w=(Int){qpow(G,(mod1-1)/limit,mod1),qpow(G,(mod2-1)/limit,mod2),qpow(G,(mod3-1)/limit,mod3)};
        Wn[0]=Int(1);
        for(int i=1;i<=limit;i++) Wn[i]=Wn[i-1]*w;//只有这里一个地方取=
        ilim=(Int){inv(limit,mod1),inv(limit,mod2),inv(limit,mod3)};
    }
    inline void NTT(Int *A,int type){
        for(int i=0;i<limit;i++)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int len=1;len<limit;len<<=1){
            int t=(limit/len)>>1;
            for(int i=0;i<limit;i+=(len<<1))
                for(int j=0;j<len;j++){
                    Int w=(type==1)?Wn[t*j]:Wn[limit-t*j];
                    Int x=A[i+j],y=w*A[i+len+j];
                    A[i+j]=x+y,A[i+len+j]=x-y;
                }
        }
        if(type==-1){
            for(int i=0;i<limit;i++) A[i]=A[i]*ilim;
        }
    }
}using namespace N_T_T;

int n,m;

int main(){
    n=read(),m=read(),mod=read();
    for(int i=0;i<=n;i++) A[i]=Int(read()%mod);
    for(int i=0;i<=m;i++) B[i]=Int(read()%mod);
    init(n+m+2);
    NTT(A,1);NTT(B,1);
    for(int i=0;i<limit;i++) A[i]=A[i]*B[i];
    NTT(A,-1);//只要做三次
    for(int i=0;i<=n+m;i++)
        printf("%d ",A[i].get());
}