[Explanations] Luogu P4245 [template] NTT any modulus

Tri-mode NTTwill not. . .

Are 0202 years, there are people writing tri-mode NTT ah. . .

---

Write to tell a good practice point of it:

First, take a threshold \ (W \) , then each of the coefficients of a polynomial written \ (aw + c (c < w) \) form, in other words the polynomial \ (f (x) \) written in two added in the form of a polynomial:
\ [F (X) = wf_0 (X) F_1 + (X) \]
Thus in this question take the \ (W = 2 ^ {15 } \) can be avoided burst long longof.

Multiply the word is
\ [f \ cdot g = (
w f_0 + f_1) (wg_0 + g_1) = (f_0 g_0) w ^ 2 + (f_0g_1 + f_1g_0) w + f_1g_1 \] so we only count \ (f_0g_0, ( f_0g_1 + f_1g_0), f_1g_1 \) just fine, respectively \ (FFT \) count. (This stuff is like asking \ (MTT \) , wonderful ~ ah)

So do \ (7 \) times \ (the FFT \) , as if you could do \ (4 \) times, no.

This also feeling ran very fast.

\(Code\)

#include <bits/stdc++.h>
using namespace std;
typedef long double db;
typedef long long ll;
const db PI=acos(-1.0);
const int N=3e5+10;
struct cpl{
    db x,y;
    cpl operator + (cpl k1)const{return (cpl){x+k1.x,y+k1.y};}
    cpl operator - (cpl k1)const{return (cpl){x-k1.x,y-k1.y};}
    cpl operator * (cpl k1)const{return (cpl){x*k1.x-y*k1.y,x*k1.y+y*k1.x};}
};
int rev[N];
void fft(cpl *f,int n,int k1){
    for (int i=0;i<n;i++)
        if (rev[i]<i)swap(f[i],f[rev[i]]);
    for (int len=2;len<=n;len<<=1){
        cpl wn=(cpl){cos(2*PI/len),k1*sin(2*PI/len)};
        for (int i=0;i<n;i+=len){
            cpl w=(cpl){1,0};
            for (int j=i;j<i+(len>>1);j++){
                cpl tmp=w*f[j+(len>>1)];
                f[j+(len>>1)]=f[j]-tmp;
                f[j]=f[j]+tmp;
                w=w*wn;
            }
        }
    }
}
cpl f[2][N],g[2][N],ans[3][N];
#define normal(x) (((ll)(x/limit+0.5)%mod+mod)%mod)
void mtt(int *a,int n,int *b,int m,int mod){
    int limit=1; while (limit<=n+m)limit<<=1;
    for (int i=0;i<limit;i++) rev[i]=rev[i>>1]>>1|((i&1)?limit>>1:0);
    for (int i=0;i<limit;i++){
        f[0][i].x=a[i]>>15;f[1][i].x=a[i]&0x7fff;
        g[0][i].x=b[i]>>15;g[1][i].x=b[i]&0x7fff;
    }
    fft(f[0],limit,1),fft(f[1],limit,1);
    fft(g[0],limit,1),fft(g[1],limit,1);
    for (int i=0;i<limit;i++){
        ans[0][i]=f[0][i]*g[0][i];
        ans[1][i]=f[0][i]*g[1][i]+f[1][i]*g[0][i];
        ans[2][i]=f[1][i]*g[1][i];
    }
    fft(ans[0],limit,-1),fft(ans[1],limit,-1),fft(ans[2],limit,-1);
    for (int i=0;i<=n+m;i++){
        ll k1=(normal(ans[0][i].x)<<30ll)%mod;
        ll k2=(normal(ans[1][i].x)<<15ll)%mod;
        ll k3=normal(ans[2][i].x)%mod;
        printf("%d ",((k1+k2)%mod+k3)%mod);
    }
}
int n,m,a[N],b[N],mod;
int main(){
    scanf("%d%d%d",&n,&m,&mod);
    for (int i=0;i<=n;i++)scanf("%d",&a[i]);
    for (int i=0;i<=m;i++)scanf("%d",&b[i]);
    mtt(a,n,b,m,mod);
    return 0;
}