FFT模板

整理了一下，自己写了一下模板

const double PI=acos(-1.0);
struct complex
{double r,i;complex(double _r=0,double _i=0):r(_r),i(_i){}complex operator +(const complex &b) {return complex(r+b.r,i+b.i);}complex operator -(const complex &b) {return complex(r-b.r,i-b.i);}complex operator *(const complex &b) {return complex(r*b.r-i*b.i,r*b.i+i*b.r);}
}A[MAXN],B[MAXN];void change(complex y[],int len)
{int i,j,k;for(i = 1, j = len/2;i < len-1;i++){if(i < j)swap(y[i],y[j]);k = len/2;while( j >= k){j -= k;k /= 2;}if(j < k)j += k;}
}void fft(complex y[],int len,int on)
{change(y,len);for(int h = 2;h <= len;h <<= 1){complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));for(int j = 0;j < len;j += h){complex w(1,0);for(int k = j;k < j+h/2;k++){complex u = y[k];complex t = w*y[k+h/2];y[k] = u+t;y[k+h/2] = u-t;w = w*wn;}}}if(on == -1)for(int i = 0;i < len;i++)y[i].r /= len;
}void FFT(int a[],int la,int b[],int lb)//la,lb分别是a,b数组的最高位+1 
{int len=1; while(len<la+lb) len<<=1;for(int i=0;i<la;++i) A[i]=complex(a[i],0);for(int i=la;i<len;++i) A[i]=complex(0,0);for(int i=0;i<lb;++i) B[i]=complex(b[i],0);for(int i=lb;i<len;++i) B[i]=complex(0,0);fft(A,len,1); fft(B,len,1);for(int i=0;i<len;++i) A[i]=A[i]*B[i];fft(A,len,-1);
}