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(-);
 struct complex{
     double a,b;
     complex(double _a=,double _b=){
         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[],b[];
 int n,m,bit,bitnum=,rev[pw()];
 void getrev(int l){//Reverse
     for(int i=;i<pw(l);i++){
         rev[i]=(rev[i>>]>>)|((i&)<<(l-));
     }
 }
 void FFT(complex *s,int op){
     for(int i=;i<bit;i++)if(i<rev[i])swap(s[i],s[rev[i]]);
     for(int i=;i<bit;i<<=){
         complex w(cos(pi/i),op*sin(pi/i));
         for(int p=i<<,j=;j<bit;j+=p){//Butterfly
             complex wk(,);
             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==-){
         for(int i=;i<=bit;i++){
             s[i]=s[i]/(double)bit;
         }
     }
 }
 int main(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)scanf("%lf",&a[i].a);
     for(int i=;i<=m;i++)scanf("%lf",&b[i].a);
     m+=n;
     for(bit=;bit<=m;bit<<=)bitnum++;
     getrev(bitnum);
     FFT(a,);
     FFT(b,);
     for(int i=;i<=bit;i++)a[i]=a[i]*b[i];
     FFT(a,-);
     for(int i=;i<=m;i++)printf("%d ",(int)(a[i].a+0.5));
     return ;
 }

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=,P=,g=;//或P=1004535809
 int n,m,bit,bitnum=,a[N+],b[N+],rev[N+];
 void getrev(int l){
     for(int i=;i<pw(l);i++){
         rev[i]=(rev[i>>]>>)|((i&)<<(l-));
     }
 }
 int fastpow(int a,int b){
     int ans=;
     for(;b;b>>=,a=1LL*a*a%P){
         if(b&)ans=1LL*ans*a%P;
     }
     return ans;
 }
 void NTT(int *s,int op){
     for(int i=;i<bit;i++)if(i<rev[i])swap(s[i],s[rev[i]]);
     for(int i=;i<bit;i<<=){
         int w=fastpow(g,(P-)/(i<<));
         for(int p=i<<,j=;j<bit;j+=p){
             int wk=;
             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==-){
         reverse(s+,s+bit);
         int inv=fastpow(bit,P-);
         for(int i=;i<bit;i++)a[i]=1LL*a[i]*inv%P;
     }
 }
 int main(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)scanf("%d",&a[i]);
     for(int i=;i<=m;i++)scanf("%d",&b[i]);
     m+=n;
     for(bit=;bit<=m;bit<<=)bitnum++;
     getrev(bitnum);
     NTT(a,);
     NTT(b,);
     for(int i=;i<bit;i++)a[i]=1LL*a[i]*b[i]%P;
     NTT(a,-);
     for(int i=m;i>=;i--)printf("%d ",a[i]);
     return ;
 }