## bzoj 2194 快速傅立叶之二

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define db double
using namespace std;
const int N=1e5+5,M=N<<2; const db pi=acos(-1);
int n,r[M],len;
struct cpl{db x,y;}a[M],b[M],I;
cpl operator+ (cpl a,cpl b){return (cpl){a.x+b.x,a.y+b.y};}
cpl operator- (cpl a,cpl b){return (cpl){a.x-b.x,a.y-b.y};}
cpl operator* (cpl a,cpl b){return (cpl){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
int rdn()
{
int ret=0;bool fx=1;char ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')fx=0;ch=getchar();}
while(ch>='0'&&ch<='9') ret=(ret<<3)+(ret<<1)+ch-'0',ch=getchar();
return fx?ret:-ret;
}
void fft(cpl *a,bool fx)
{
for(int i=0;i<len;i++)
if(i<r[i])swap(a[i],a[r[i]]);
for(int R=2;R<=len;R<<=1)
{
int m=R>>1;
cpl Wn=(cpl){ cos(pi/m),fx?-sin(pi/m):sin(pi/m) };
for(int i=0;i<len;i+=R)
{
cpl w=I;
for(int j=0;j<m;j++,w=w*Wn)
{
cpl tmp=w*a[i+m+j];
a[i+m+j]=a[i+j]-tmp;
a[i+j]=a[i+j]+tmp;
}
}
}
}
int main()
{
n=rdn(); I.x=1;
for(int i=0;i<n;i++)
a[n-1-i].x=rdn(),b[i].x=rdn();
len=1;
for(;len<=n<<1;len<<=1);
for(int i=0;i<len;i++)
r[i]=(r[i>>1]>>1)+((i&1)?len>>1:0);
fft(a,0); fft(b,0);
for(int i=0;i<len;i++)a[i]=a[i]*b[i];
fft(a,1);
for(int i=n-1;i>=0;i--)
printf("%d\n",int(a[i].x/len+0.5));
return 0;
}

posted on 2018-11-26 15:46  Narh  阅读(103)  评论(0编辑  收藏  举报