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
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <cmath> 5 #include <set> 6 #include <map> 7 #include <list> 8 #include <queue> 9 #include <vector> 10 #include <bitset> 11 #include <iostream> 12 using namespace std; 13 14 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 15 const double pi=acos(-1); 16 17 struct Complex 18 { 19 double x,y; 20 ///otherwise create Complex(){} 21 Complex(double xx=0,double yy=0) 22 { 23 x=xx; 24 y=yy; 25 } 26 Complex operator+(const Complex &b) const 27 { 28 return Complex(x+b.x,y+b.y); 29 } 30 Complex operator-(const Complex &b) const 31 { 32 return Complex(x-b.x,y-b.y); 33 } 34 Complex operator*(const Complex &b) const 35 { 36 return Complex(x*b.x-y*b.y,x*b.y+y*b.x); 37 } 38 }a[maxn],b[maxn]; 39 40 int w=0,s,r[maxn]; 41 42 void fft(Complex *a,int inv) 43 { 44 int i,j,m; 45 for (i=0;i<s;i++) 46 if (i<r[i]) 47 swap(a[i],a[r[i]]); 48 for (m=1;m<s;m<<=1) 49 { 50 Complex temp(cos(pi/m),inv*sin(pi/m)); 51 for (i=0;i<s;i+=m*2) 52 { 53 Complex mul(1,0); 54 for (j=0;j<m;j++,mul=mul*temp) 55 { 56 Complex x=a[i+j],y=mul*a[i+j+m]; 57 a[i+j]=x+y,a[i+j+m]=x-y; 58 } 59 } 60 } 61 } 62 63 int main() 64 { 65 int n,m,i; 66 scanf("%d%d",&n,&m); 67 for (i=0;i<=n;i++) 68 scanf("%lf",&a[i].x); 69 for (i=0;i<=m;i++) 70 scanf("%lf",&b[i].x); 71 s=1; 72 while (s<=n+m) //n+m+1 73 s<<=1,w++; 74 for (i=0;i<s;i++) 75 r[i]=(r[i>>1]>>1) | ((i & 1)<<(w-1)); 76 fft(a,1); 77 fft(b,1); 78 for (i=0;i<=s;i++) 79 a[i]=a[i]*b[i]; 80 fft(a,-1); 81 for (i=0;i<=n+m;i++) 82 printf("%d ",(int)(a[i].x/s+0.5)); 83 return 0; 84 }
NTT
......
FWT
https://www.cnblogs.com/cjyyb/p/9065615.html
