把FFT模版略微改动一下
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
const int N=270000;
const double Pi=acos(-1);
int An,Bn,Cn,n,Rev[N],Step;
struct Complex{
double a,b;
Complex(){}
Complex(double sa,double sb){a=sa;b=sb;}
friend Complex operator+(Complex A,Complex B){return Complex(A.a+B.a,A.b+B.b);}
friend Complex operator-(Complex A,Complex B){return Complex(A.a-B.a,A.b-B.b);}
friend Complex operator*(Complex A,Complex B){return Complex(A.a*B.a-A.b*B.b,A.a*B.b+A.b*B.a);}
}A[N],B[N],C[N];
void FFT(Complex *x,int flag){
for(int i=0;i<n;i++)if(i<Rev[i])swap(x[i],x[Rev[i]]);
for(int k=1;k<n;k<<=1){
Complex wk=Complex(cos(Pi/k),flag*sin(Pi/k));
for(int i=0;i<n;i+=k<<1){
Complex wkj=Complex(1.0,0.0);
for(int j=0;j<k;j++){
Complex a=x[i+j],b=x[i+j+k]*wkj;
x[i+j]=a+b;
x[i+j+k]=a-b;
wkj=wkj*wk;
}
}
}
if(flag==-1)for(int i=0;i<n;i++)x[i].a/=n;
}
int main(){
freopen("2194.in","r",stdin);
freopen("2194.out","w",stdout);
scanf("%d",&An);
Bn=An;Cn=An+Bn-1;
for(int i=0;i<An;i++)scanf("%lf %lf",&A[i].a,&B[An-i].a);
for(n=1,Step=0;n<Cn;n<<=1,Step++);
for(int i=0;i<n;i++)Rev[i]=(Rev[i>>1]>>1)|((i&1)<<(Step-1));
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=Cn/2+1;i<=Cn;i++)printf("%d\n",(int)(C[i].a+0.5));
return 0;
}
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