bzoj 2194 快速傅立叶之二

题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2194

因为卷积的第 k 项是 sigma(i=0~k)a[ i ]*b[ k-i ] ,也就是角标加起来是 k 的两项求和;所以先把 a 序列翻转一下,然后发现正常卷积的第 n-1-k 项就是 c[k] 。

#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

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