bzoj3527[Zjoi2014]力 FFT

Posted 2020-07-18 wjyi

2016-06-01  21:36:44

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3527

我就是一个大傻叉 微笑脸

 1 #include<bits/stdc++.h>
 2 #define inf 1000000000
 3 #define ll long long
 4 #define N 500005
 5 using namespace std;
 6 int read(){
 7   int x=0,f=1;char ch=getchar();
 8   while(ch<\'0\'||ch>\'9\'){if(ch==\'-\')f=-1;ch=getchar();}
 9   while(ch>=\'0\'&&ch<=\'9\'){x=x*10+ch-\'0\';ch=getchar();}
10   return x*f;
11 }
12 const double Pi=acos(-1.0);
13 struct CD{
14   double x,y;
15   CD(double a=0,double b=0){x=a,y=b;}
16   friend CD operator + (CD n1,CD n2){return CD(n1.x+n2.x,n1.y+n2.y);}
17   friend CD operator - (CD n1,CD n2){return CD(n1.x-n2.x,n1.y-n2.y);}
18   friend CD operator * (CD n1,CD n2){return CD(n1.x*n2.x-n1.y*n2.y,n1.x*n2.y+n1.y*n2.x);}
19 };
20 CD a[N],b[N],c[N],d[N];
21 int n,nn,bit;
22 double q[N];
23 void FFT(CD *a,int n,int type){
24   for(int i=0,j=0;i<n;i++){
25     if(j>i)swap(a[i],a[j]);
26     int k=n;
27     while(j&(k>>=1))j&=~k;
28     j|=k;
29   }
30   for(int i=1;i<=bit;i++){
31     CD w_n(cos(2*type*Pi/(1<<i)),sin(2*type*Pi/(1<<i)));
32     for(int j=0;j<(1<<bit);j+=(1<<i)){
33       CD w(1,0);
34       for(int k=j;k<j+(1<<(i-1));k++){
35         CD tmp=a[k],tt=w*a[k+(1<<(i-1))];
36         a[k]=tmp+tt;a[k+(1<<(i-1))]=tmp-tt;
37         w=w*w_n;
38       }
39     }
40   }
41   if(type<0)for(int i=0;i<n;i++)a[i].x=a[i].x/n;
42 }
43 int main(){
44   n=read();nn=n;
45   for(int i=0;i<n;i++)scanf("%lf",&q[i]),a[i]=CD(q[i],0);
46   for(int i=1;i<n;i++)b[i].x=1.0/(double)(i*i),b[i].y=0;
47   n=2*n-1;bit=1;
48   while((1<<bit)<n)bit++;
49   n=1<<bit;
50   b[0]=CD();
51   for(int i=nn;i<n;i++)a[i]=CD(),b[i]=CD();
52    
53   FFT(a,n,1);FFT(b,n,1);
54   for(int i=0;i<n;i++)c[i]=a[i]*b[i];
55   FFT(c,n,-1);
56   for(int i=0;i<nn;i++)a[i]=CD(q[nn-i-1],0);
57   for(int i=nn;i<n;i++)a[i]=CD();
58   FFT(a,n,1);
59   for(int i=0;i<n;i++)d[i]=a[i]*b[i];
60   FFT(d,n,-1);
61    
62   for(int i=0;i<nn;i++)c[i].x-=d[nn-i-1].x;
63   for(int i=0;i<nn;i++)printf("%.5lf\\n",c[i].x);
64   return 0;
65 }
66 

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