快速傅里叶变换

study from：

http://www.cnblogs.com/zwfymqz/p/8244902.html

https://blog.csdn.net/enjoy_pascal/article/details/81478582

luogu3803

https://www.luogu.org/problemnew/show/P3803

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <vector>
#include <bitset>
#include <iostream>
using namespace std;

const int maxn=(1e6+10)*4;///相乘，最多2e6位，同时要加到min 2^k>2e6，乘2即可
const double pi=acos(-1);

struct Complex
{
    double x,y;
    ///otherwise create Complex(){}
    Complex(double xx=0,double yy=0)
    {
        x=xx;
        y=yy;
    }
    Complex operator+(const Complex &b) const
    {
        return Complex(x+b.x,y+b.y);
    }
    Complex operator-(const Complex &b) const
    {
        return Complex(x-b.x,y-b.y);
    }
    Complex operator*(const Complex &b) const
    {
        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
}a[maxn],b[maxn];

int w=0,s,r[maxn];

void fft(Complex *a,int inv)
{
    int i,j,m;
    for (i=0;i<s;i++)
        if (i<r[i])
            swap(a[i],a[r[i]]);
    for (m=1;m<s;m<<=1)
    {
        Complex temp(cos(pi/m),inv*sin(pi/m));
        for (i=0;i<s;i+=m*2)
        {
            Complex mul(1,0);
            for (j=0;j<m;j++,mul=mul*temp)
            {
                Complex x=a[i+j],y=mul*a[i+j+m];
                a[i+j]=x+y,a[i+j+m]=x-y;
            }
        }
    }
}

int main()
{
    int n,m,i;
    scanf("%d%d",&n,&m);
    for (i=0;i<=n;i++)
        scanf("%lf",&a[i].x);
    for (i=0;i<=m;i++)
        scanf("%lf",&b[i].x);
    s=1;
    while (s<=n+m)  //n+m+1
        s<<=1,w++;
    for (i=0;i<s;i++)
        r[i]=(r[i>>1]>>1) | ((i & 1)<<(w-1));
    fft(a,1);
    fft(b,1);
    for (i=0;i<=s;i++)
        a[i]=a[i]*b[i];
    fft(a,-1);
    for (i=0;i<=n+m;i++)
        printf("%d ",(int)(a[i].x/s+0.5));
    return 0;
}

NTT
......

FWT

https://www.cnblogs.com/cjyyb/p/9065615.html