bzoj 2179 FFT快速傅立叶 —— FFT

题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2179

默写板子,注释的是忘记的地方。

代码如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef double db;
int const xn=(1<<17);
db const Pi=acos(-1.0);
int n,lim,l,rev[xn],ans[xn];
struct com{db x,y;}a[xn],b[xn];
com operator + (com a,com b){return (com){a.x+b.x,a.y+b.y};}
com operator - (com a,com b){return (com){a.x-b.x,a.y-b.y};}
com operator * (com a,com b){return (com){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
int rd()
{
  int ret=0,f=1; char ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-')f=0; ch=getchar();}
  while(ch>='0'&&ch<='9')ret=(ret<<3)+(ret<<1)+ch-'0',ch=getchar();
  return f?ret:-ret;
}
void fft(com *a,int tp)
{
  for(int i=0;i<lim;i++)
    if(i<rev[i])swap(a[i],a[rev[i]]);
  for(int mid=1;mid<lim;mid<<=1)//mid<<=1
    {
      com wn=(com){cos(Pi/mid),tp*sin(Pi/mid)};
      for(int j=0,len=(mid<<1);j<lim;j+=len)
    {
      com w=(com){1,0};
      for(int k=0;k<mid;k++,w=w*wn)//<
        {
          com x=a[j+k],y=w*a[j+mid+k];
          a[j+k]=x+y; 
          a[j+mid+k]=x-y;
        }
    }
    }
}
int main()
{
  n=rd()-1;
  for(int i=0,t;i<=n;i++)scanf("%1d",&t),a[i].x=t;
  for(int i=0,t;i<=n;i++)scanf("%1d",&t),b[i].x=t;
  lim=1;
  while(lim<=n+n)lim<<=1,l++;
  for(int i=0;i<lim;i++)
    rev[i]=((rev[i>>1]>>1)|((i&1)<<(l-1)));
  fft(a,1); fft(b,1);
  for(int i=0;i<lim;i++)a[i]=a[i]*b[i];//<lim
  fft(a,-1);
  for(int i=0;i<=n+n;i++)ans[i]=(int)(a[i].x/lim+0.5);
  for(int i=n+n;i;i--)
      if(ans[i]>=10)ans[i-1]+=ans[i]/10,ans[i]%=10;
  int i=0;
  while(!ans[i])i++;
  while(i<=n+n)printf("%d",ans[i]),i++; puts("");
  return 0;
}

 

posted @ 2018-11-26 15:01  Zinn  阅读(120)  评论(0编辑  收藏  举报