P4245 【模板】任意模数NTT
作者:互联网
#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());
}
标签:mod2,mod3,Int,NTT,模数,int,inline,P4245,mod 来源: https://www.cnblogs.com/lizehon/p/10501725.html