bzoj3527 [Zjoi2014]力 FFT

[Zjoi2014]力

Time Limit: 30 Sec Memory Limit: 256 MBSec Special Judge
Submit: 2719 Solved: 1656
[Submit][Status][Discuss]

Description

给出n个数qi，给出Fj的定义如下：

令Ei=Fi/qi，求Ei.

Input

第一行一个整数n。

接下来n行每行输入一个数，第i行表示qi。

n≤100000，0<qi<1000000000

Output

n行，第i行输出Ei。与标准答案误差不超过1e-2即可。

Sample Input

5
4006373.885184
15375036.435759
1717456.469144
8514941.004912
1410681.345880

Sample Output

-16838672.693
3439.793
7509018.566
4595686.886
10903040.872

HINT

https://www.cnblogs.com/iwtwiioi/p/4126284.html

 1 #pragma GCC optimize(2)
 2 #pragma G++ optimize(2)
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<algorithm>
 8 
 9 #define N 300007
10 #define pi acos(-1)
11 using namespace std;
12 inline int read()
13 {
14     int x=0,f=1;char ch=getchar();
15     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
16     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
17     return x*f;
18 }
19 
20 int n,m,L;
21 int rev[N];
22 struct comp
23 {
24     double r,v;
25     inline comp operator+(comp const &a){return (comp){r+a.r,v+a.v};}
26     inline comp operator-(comp const &a){return (comp){r-a.r,v-a.v};}
27     inline comp operator*(comp const &a){return (comp){r*a.r-v*a.v,r*a.v+v*a.r};}
28 }f[N],_f[N],g[N],e1[N],e2[N];
29 
30 void FFT(comp *a,int f)
31 {
32     for (int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
33     for (int i=1;i<n;i<<=1)
34     {
35         comp wn=(comp){cos(pi/i),f*sin(pi/i)};
36         for (int j=0;j<n;j+=(i<<1))
37         {
38             comp w=(comp){1,0};
39             for (int k=0;k<i;k++,w=w*wn)
40             {
41                 comp x=a[j+k],y=w*a[j+k+i];
42                 a[j+k]=x+y,a[j+k+i]=x-y;
43             }
44         }
45     }
46     if(f==-1)for (int i=0;i<n;i++)a[i].r/=n;
47 }
48 int main()
49 {
50     n=read()-1;
51     for (int i=0;i<=n;i++)
52     {
53         double x;scanf("%lf",&x);
54         f[i].r=_f[n-i].r=x;
55     }
56     for (int i=1;i<=n;i++)g[i].r=(1.0/i/i);
57     m=2*n;for (n=1;n<=m;n<<=1,L++);if(L)L--;
58     for (int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
59     FFT(f,1),FFT(_f,1),FFT(g,1);
60     for (int i=0;i<n;i++)e1[i]=f[i]*g[i],e2[i]=_f[i]*g[i];
61     FFT(e1,-1),FFT(e2,-1);
62     for (int i=0;i<=m/2;i++)
63         printf("%.7lf
",e1[i].r-e2[m/2-i].r);
64 }