题链:
http://uoj.ac/problem/34
题解:
FFT入门题。
(终于接触到迷一样的FFT了)
初学者在对复数和单位根有简单了解的基础上,可以直接看《再探快速傅里叶变换》(毛啸)。
(主要用于求两个序列的卷积)
代码:
递归版:
#include<bits/stdc++.h>
#define MAXN 300000
using namespace std;
const double Pi=acos(-1);
struct Z{
double real,image;
Z(double _real=0,double _image=0):real(_real),image(_image){}
Z operator - ()const {return Z(-real,-image);}
friend Z operator + (const Z &A,const Z &B){return Z(A.real+B.real,A.image+B.image);};
friend Z operator - (const Z &A,const Z &B){return A+(-B);}
friend Z operator * (const Z &A,const Z &B){return Z(A.real*B.real-A.image*B.image,A.image*B.real+A.real*B.image);}
};
void FFT(int n,Z *Y,int sn){
if(n==1) return;
Z L[n>>1],R[n>>1];
for(int k=0;k<n;k+=2)
L[k>>1]=Y[k],R[k>>1]=Y[k+1];
FFT(n/2,L,sn); FFT(n/2,R,sn);
Z dw=Z(cos(2*Pi/n),sin(sn*2*Pi/n)),w=Z(1,0),tmp;
for(int k=0;k<n/2;k++,w=w*dw)
tmp=w*R[k],Y[k]=L[k]+tmp,Y[k+n/2]=L[k]-tmp;
}
int main(){
static Z A[MAXN],B[MAXN];
int n,m; scanf("%d%d",&n,&m); n++; m++;
for(int i=0,x;i<n;i++) scanf("%d",&x),A[i]=Z(x,0);
for(int i=0,x;i<m;i++) scanf("%d",&x),B[i]=Z(x,0);
m=n+m-1; for(n=1;n<m;n<<=1);
FFT(n,A,1); FFT(n,B,1);
for(int i=0;i<n;i++) A[i]=A[i]*B[i];
FFT(n,A,-1);
for(int i=0;i<m;i++) printf("%d ",(int)(A[i].real/n+0.5));
return 0;
}
非递归版:
#include<bits/stdc++.h>
#define MAXN 300000
using namespace std;
const double Pi=acos(-1);
struct Z{
double real,image;
Z(double _real=0,double _image=0):real(_real),image(_image){}
Z operator - ()const {return Z(-real,-image);}
friend Z operator + (const Z &A,const Z &B){return Z(A.real+B.real,A.image+B.image);};
friend Z operator - (const Z &A,const Z &B){return A+(-B);}
friend Z operator * (const Z &A,const Z &B){return Z(A.real*B.real-A.image*B.image,A.image*B.real+A.real*B.image);}
};
int order[MAXN];
void FFT(int n,Z *Y,int sn){
for(int i=0;i<n;i++) if(i<order[i]) swap(Y[i],Y[order[i]]);
for(int d=2;d<=n;d<<=1){
Z dw=Z(cos(2*Pi/d),sin(sn*2*Pi/d)),w,tmp;
for(int i=0;w=Z(1,0),i<n;i+=d)
for(int k=i;k<i+d/2;k++,w=w*dw)
tmp=w*Y[k+d/2],Y[k+d/2]=Y[k]-tmp,Y[k]=Y[k]+tmp;
}
}
int main(){
static Z A[MAXN],B[MAXN];
int n,m,len; scanf("%d%d",&n,&m); n++; m++;
for(int i=0,x;i<n;i++) scanf("%d",&x),A[i]=Z(x,0);
for(int i=0,x;i<m;i++) scanf("%d",&x),B[i]=Z(x,0);
m=n+m-1; for(len=0,n=1;n<m;n<<=1) len++;
for(int i=1;i<n;i++) order[i]=(order[i>>1]>>1)|((i&1)<<(len-1));
FFT(n,A,1); FFT(n,B,1);
for(int i=0;i<n;i++) A[i]=A[i]*B[i];
FFT(n,A,-1);
for(int i=0;i<m;i++) printf("%d ",(int)(A[i].real/n+0.5));
return 0;
}
转载于:https://www.cnblogs.com/zj75211/p/8341072.html