三模数NTT模板

求两个多项式的卷积对任意数p取模

两个好记的FNT模数：

5*2^25+1

7*2^26+1

原根都为3

 //Achen
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 #include<cstdio>
 #include<queue>
 #include<cmath>
 #include<set>
 #include<map>
 #define Formylove return 0
 #define For(i,a,b) for(int i=(a);i<=(b);i++)
 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
 const int N=,g=;
 typedef long long LL;
 typedef double db;
 typedef long double LD;
 using namespace std;
 int n,m,mod;
 LL a[][N],b[][N],p[]={,,},gi[]={,,};
 LL ans[N];

 template<typename T>void read(T &x)  {
     char ch=getchar(); x=; T f=;
     while(ch!='-'&&(ch<''||ch>'')) ch=getchar();
     if(ch=='-') f=-,ch=getchar();
     for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-''; x*=f;
 }

 LL ksc(LL a,LL b,LL p) {
     LL tp=a*b-(LL)((LD)a/p*b+1.0e-8)*p;
     return tp<?tp+p:tp%p;
 }

 LL ksm(LL a,LL b,LL p) {
     LL rs=,bs=a%p;
     while(b) {
         if(b&) rs=ksc(rs,bs,p);
         bs=ksc(bs,bs,p);
         b>>=;
     }
     return rs;
 }

 int l,rev[N];
 void FFT(int n,LL a[],int f,int p,int gi) {
     For(i,,n-) if(i<rev[i]) swap(a[i],a[rev[i]]);
     for(int i=;i<n;i<<=) {
         LL wi=ksm((f==)?g:gi,(p-)/(i<<),p);
         for(int j=,pp=(i<<);j<n;j+=pp) {
             LL w=;
             for(int k=;k<i;k++,w=w*wi%p) {
                 LL x=a[j+k],y=w*a[j+k+i]%p;
                 a[j+k]=(x+y)%p; a[j+k+i]=(x-y+p)%p;
             }
         }
     }
     if(f==-) {
         LL inv=ksm(n,p-,p);
         For(i,,n) a[i]=a[i]*inv%p;
     }
 }

 int main() {
 #ifdef ANS
     freopen(".in","r",stdin);
     freopen(".out","w",stdout);
 #endif
     read(n); read(m); read(mod);
     For(i,,n) { read(a[][i]); a[][i]=a[][i]=a[][i]; }
     For(i,,m) { read(b[][i]); b[][i]=b[][i]=b[][i]; }
     m+=n;
     for(n=;n<=m;n<<=) l++;
     For(i,,n) rev[i]=(rev[i>>]>>)|((i&)<<(l-));
     For(i,,) {
         FFT(n,a[i],,p[i],gi[i]);
         FFT(n,b[i],,p[i],gi[i]);
         For(j,,n) a[i][j]=a[i][j]*b[i][j]%p[i];
         FFT(n,a[i],-,p[i],gi[i]);
     }
     LL p1=p[],p2=p[],p3=p[];
     For(i,,m) {
         LL b1=a[][i],b2=a[][i],b3=a[][i],b4;
         b4=(ksc(ksc(b1,ksm(p2,p1-,p1),p1*p2),p2,p1*p2)+ksc(ksc(b2,ksm(p1,p2-,p2),p1*p2),p1,p1*p2))%(p1*p2);
         LL k=((b3%p3-b4%p3+p3)%p3)*ksm(p1*p2,p3-,p3)%p3;
         ans[i]=(b4%mod+k*p1%mod*p2%mod)%mod;
     }
     For(i,,m) printf("%lld ",ans[i]); puts("");
     Formylove;
 }