FFT 快速傅里叶变换

http://blog.miskcoo.com/2015/04/polynomial-multiplication-and-fast-fourier-transform#i-10
唔看了算导和一个讲的很清楚的博客
我只是做了下笔记，尽量简洁一点写吧。

这样复杂度就是

T(n)=2T(n/2)+O(n)=O(n log n)

好像发现了惊天大秘密QwQ 它们的位置是颠倒的诶！
怎么计算rev数组呢？

for (int i=1;ii++)
    rev[i]=rev[i>>1]>>1 | ((i&1) << l-1);

再看一下迭代的过程

void fft(complex *a){
    for (int i=1;iif (ifor (int len=2;len<=n;len<<=1){
        complex wn(cos(2*PI/len),sin(2*PI/len));
        for (int j=0;jlen){
            complex w(1,0),x;
            for (int k=0;k<(len>>1);k++,w*=wn)
                x=w*a[j+k+(len>>1)],
                a[j+k+(len>>1)]=a[j+k]-x,
                a[j+k]+=x;
        }
    }
}

---

最后贴一下代码（UOJ 34 模板题）

#include 
#define N 300010
#define PI acos(-1)
using namespace std;
int n,m,l,rev[N];
complex<double> a[N],b[N],c[N];

template <class Aqua>
inline void read(Aqua &s){
    s=0; char c=getchar();
    while (!isdigit(c)) c=getchar();
    while (isdigit(c)) s=s*10+c-'0',c=getchar();
}

void pre(){
    for (m+=n,n=1,l=0;n<=m;n<<=1,l++);
    for (int i=1;i>1]>>1 | ((i&1) << l-1);
}

void fft(complex<double> *a,int f){
    for (int i=1;iif (ifor (int len=2;len<=n;len<<=1){
        complex<double> wn(cos(2*PI/len),f*sin(2*PI/len));
        for (int j=0;jcomplex<double> w(1,0),x;
            for (int k=0;k<(len>>1);k++,w*=wn)
                x=w*a[j+k+(len>>1)],
                a[j+k+(len>>1)]=a[j+k]-x,
                a[j+k]+=x;
        }
    }
}

int main(){
    read(n),read(m);
    for (int i=0;i<=n;i++)
        read(a[i].real());
    for (int i=0;i<=m;i++)
        read(b[i].real());
    pre();
    fft(a,1); fft(b,1);
    for (int i=0;i<=n;i++)
        c[i]=a[i]*b[i];
    fft(c,-1);
    for (int i=0;iprintf("%d ",(int)(c[i].real()/n+0.5));
    printf("%d\n",(int)(c[m].real()/n+0.5));
    return 0;
}

讲完啦
运用下次再写吧QwQ

UPD by 2018.2.9
附上NTT模板（UOJ34）

#include 
#define N 300010
#define P 998244353
#define ll long long
using namespace std;
int n,m,l,rev[N],a[N],b[N],c[N];

void pre(){
  for (m+=n,l=0,n=1;n<=m;l++,n<<=1);
  for (int i=1;i>1]>>1 | ((i&1) << l-1);
}

ll mpow(ll a,ll b){
  ll s=1;
  for (;b;b>>=1,a=a*a%P)
    if (b&1) s=s*a%P;
  return s;
}

void fft(int *a,int f){
  for (int i=1;iif (ifor (int len=2;len<=n;len<<=1){
    int wn=mpow(3,(P-1)/len),x;
    if (f==-1) wn=mpow(wn,P-2);
    for (int j=0;jint w=1;
        for (int k=0;k<(len>>1);k++,w=(ll)w*wn%P){
          x=(ll)a[j+k+(len>>1)]*w%P;
          a[j+k+(len>>1)]=(ll)(a[j+k]-x+P)%P;
          a[j+k]=(a[j+k]+x)%P;
        }
    }
  }
  if (f==-1)
     for (int i=0;i*mpow(n,P-2)%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]);
  pre();
  fft(a,1); fft(b,1);
  for (int i=0;i*b[i]%P;
  fft(c,-1);
  for (int i=0;i<m;i++)
    printf("%d ",c[i]);
  printf("%d\n",c[m]);
  return 0;
}