FFT&NTT学习笔记

具体原理就不讲了qwq，毕竟证明我也不太懂

FFT（快速傅立叶变换）&NTT（快速数论变换）

FFT

//求多项式乘积
//要求多项式A和多项式B的积多项式C
//具体操作就是
//DFT(A),DFT(B)->暴力乘积->拉格朗日插值(即IDFT(C))->C
//其中DFT表示离散傅里叶变换
//通俗的来说就是用点值表示多项式
//使用神秘单位复数根将时间复杂度降至O(nlogn)
//ps:但是常数巨大
//pps:应用非常广泛，非常多题目都要fft or ntt优化，板子一定要背熟 
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define pw(n) (1<<n)
using namespace std;
const double pi=acos(-1);
struct complex{
    double a,b;
    complex(double _a=0,double _b=0){
        a=_a;
        b=_b;
    }
    friend complex operator +(complex x,complex y){return complex(x.a+y.a,x.b+y.b);}
    friend complex operator -(complex x,complex y){return complex(x.a-y.a,x.b-y.b);}
    friend complex operator *(complex x,complex y){return complex(x.a*y.a-x.b*y.b,x.a*y.b+x.b*y.a);}
    friend complex operator *(complex x,double y){return complex(x.a*y,x.b*y);}
    friend complex operator /(complex x,double y){return complex(x.a/y,x.b/y);}
}a[100001],b[100001];
int n,m,bit,bitnum=0,rev[pw(20)];
void getrev(int l){//Reverse
    for(int i=0;i<pw(l);i++){
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
    }
}
void FFT(complex *s,int op){
    for(int i=0;i<bit;i++)if(i<rev[i])swap(s[i],s[rev[i]]);
    for(int i=1;i<bit;i<<=1){
        complex w(cos(pi/i),op*sin(pi/i));
        for(int p=i<<1,j=0;j<bit;j+=p){//Butterfly
            complex wk(1,0);
            for(int k=j;k<i+j;k++,wk=wk*w){
                complex x=s[k],y=wk*s[k+i];
                s[k]=x+y;
                s[k+i]=x-y;
            }
        }
    }
    if(op==-1){
        for(int i=0;i<=bit;i++){
            s[i]=s[i]/(double)bit;
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;i++)scanf("%lf",&a[i].a);
    for(int i=0;i<=m;i++)scanf("%lf",&b[i].a);
    m+=n;
    for(bit=1;bit<=m;bit<<=1)bitnum++;
    getrev(bitnum);
    FFT(a,1);
    FFT(b,1);
    for(int i=0;i<=bit;i++)a[i]=a[i]*b[i];
    FFT(a,-1);
    for(int i=0;i<=m;i++)printf("%d ",(int)(a[i].a+0.5));
    return 0;
}

NTT

//费马数数论变换
//大家觉得998244353好还是1004535809好？^_^ 
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define pw(n) (1<<n)
using namespace std;
const int N=262144,P=998244353,g=3;//或P=1004535809
int n,m,bit,bitnum=0,a[N+5],b[N+5],rev[N+5];
void getrev(int l){
    for(int i=0;i<pw(l);i++){
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
    }
}
int fastpow(int a,int b){
    int ans=1;
    for(;b;b>>=1,a=1LL*a*a%P){
        if(b&1)ans=1LL*ans*a%P;
    }
    return ans;
}
void NTT(int *s,int op){
    for(int i=0;i<bit;i++)if(i<rev[i])swap(s[i],s[rev[i]]);
    for(int i=1;i<bit;i<<=1){
        int w=fastpow(g,(P-1)/(i<<1));
        for(int p=i<<1,j=0;j<bit;j+=p){
            int wk=1;
            for(int k=j;k<i+j;k++,wk=1LL*wk*w%P){
                int x=s[k],y=1LL*s[k+i]*wk%P;
                s[k]=(x+y)%P;
                s[k+i]=(x-y+P)%P;
            }
        }
    }
    if(op==-1){
        reverse(s+1,s+bit);
        int inv=fastpow(bit,P-2);
        for(int i=0;i<bit;i++)a[i]=1LL*a[i]*inv%P;
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;i++)scanf("%d",&a[i]);
    for(int i=0;i<=m;i++)scanf("%d",&b[i]);
    m+=n;
    for(bit=1;bit<=m;bit<<=1)bitnum++;
    getrev(bitnum);
    NTT(a,1);
    NTT(b,1);
    for(int i=0;i<bit;i++)a[i]=1LL*a[i]*b[i]%P;
    NTT(a,-1);
    for(int i=m;i>=0;i--)printf("%d ",a[i]);
    return 0;
}