Luogu P3803 【模板】多项式乘法 (FFT) (NTT)

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Luogu P3803 【模板】多项式乘法(FFT)

　　

FFT##

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
const int MAXN=5e6+5;
const double Pi=acos(-1.0);
struct complex
{
    double x,y;
    complex (double xx=0,double yy=0) { x=xx,y=yy; }
}a[MAXN],b[MAXN];
complex operator + (complex a,complex b){return complex(a.x+b.x , a.y+b.y);}
complex operator - (complex a,complex b){return complex(a.x-b.x , a.y-b.y);}
complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);} 
int n,m,limit,len;
int p[MAXN];
inline int read()
{
    int x=0,f=1; char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void FFT(complex *A,int type)
{
    for (int i=0; i<limit; i++) 
        if (i<p[i]) swap(A[i],A[p[i]]);
    for (int l=1; l<limit; l<<=1)
    {
        complex wn(cos(Pi/l),type*sin(Pi/l));
        for (int i=0; i<limit; i+=(l<<1))
        {
            complex w(1,0);
            for (int j=0; j<l; j++,w=w*wn)
            {
                complex t=w*A[i+j+l];
                A[i+j+l]=A[i+j]-t;
                A[i+j]=A[i+j]+t;
            }
        }
    }
}
int main()
{
    n=read(); m=read(); 
    for (int i=0; i<=n; i++) a[i].x=read();
    for (int i=0; i<=m; i++) b[i].x=read();
    
    limit=1;
    while (limit<n+m+1) limit<<=1,len++;
    for (int i=0; i<limit; i++)
        p[i]=(p[i>>1]>>1)|((i&1)<<(len-1));
    
    FFT(a,1); FFT(b,1);
    for (int i=0; i<limit; i++) a[i]=a[i]*b[i];
    FFT(a,-1);
    for (int i=0; i<=n+m; i++)
        printf("%d ",(int)(a[i].x/limit+0.5));
    printf("\n");
    return 0;
}

　　
　　
　　

NTT##

#include<cstdio>
#include<iostream>
using namespace std;
const int MAXN=5e6+5;
const int mo=998244353;
const int g=3;
int n,m,limit,len,invn;
int a[MAXN],b[MAXN],p[MAXN];
inline int read()
{
    int x=0,f=1; char ch=getchar(); 
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int power(int a,int b)
{
    int res=1;
    while (b)
    {
        if (b&1) res=1ll*res*a%mo;
        a=1ll*a*a%mo;
        b>>=1;
    }
    return res;
}
void NTT(int *A,int type)
{
    for (int i=0; i<limit; i++) 
        if (i<p[i]) swap(A[i],A[p[i]]);
    for (int l=1; l<limit; l<<=1)
    {
        int wn=power(g,(mo-1)/(l<<1));
        if (type==-1) wn=power(wn,mo-2);
        for (int i=0; i<limit; i+=(l<<1))
        {
            int w=1;
            for (int j=0; j<l; j++,w=1ll*w*wn%mo)
            {
                int t=1ll*w*A[i+j+l]%mo;
                A[i+j+l]=(A[i+j]-t+mo)%mo;
                A[i+j]=(A[i+j]+t)%mo;
            }
        }
    }
}
int main()
{
    n=read(); m=read(); 
    for (int i=0; i<=n; i++) a[i]=read();
    for (int i=0; i<=m; i++) b[i]=read();
    
    limit=1;
    while (limit<n+m+1) limit<<=1,len++;
    for (int i=0; i<limit; i++)
        p[i]=(p[i>>1]>>1)|((i&1)<<(len-1));
    
    NTT(a,1); NTT(b,1);
    for (int i=0; i<limit; i++) a[i]=1ll*a[i]*b[i]%mo;
    NTT(a,-1);
    invn=power(limit,mo-2);
    for (int i=0; i<=n+m; i++)
        printf("%lld ",1ll*a[i]*invn%mo);
    printf("\n");
    return 0;
}