bzoj3527 [Zjoi2014]力 FFT

 [Zjoi2014]力

Time Limit: 30 Sec  Memory Limit: 256 MBSec  Special Judge
Submit: 2719  Solved: 1656
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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

#pragma GCC optimize(2)
#pragma G++ optimize(2)
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdio>
#include<algorithm>

#define N 300007
#define pi acos(-1)
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

int n,m,L;
int rev[N];
struct comp
{
    double r,v;
    inline comp operator+(comp const &a){return (comp){r+a.r,v+a.v};}
    inline comp operator-(comp const &a){return (comp){r-a.r,v-a.v};}
    inline comp operator*(comp const &a){return (comp){r*a.r-v*a.v,r*a.v+v*a.r};}
}f[N],_f[N],g[N],e1[N],e2[N];

void FFT(comp *a,int f)
{
    for (int i=0;i<n;i++)if(i<rev[i])swap(a[i],a[rev[i]]);
    for (int i=1;i<n;i<<=1)
    {
        comp wn=(comp){cos(pi/i),f*sin(pi/i)};
        for (int j=0;j<n;j+=(i<<1))
        {
            comp w=(comp){1,0};
            for (int k=0;k<i;k++,w=w*wn)
            {
                comp x=a[j+k],y=w*a[j+k+i];
                a[j+k]=x+y,a[j+k+i]=x-y;
            }
        }
    }
    if(f==-1)for (int i=0;i<n;i++)a[i].r/=n;
}
int main()
{
    n=read()-1;
    for (int i=0;i<=n;i++)
    {
        double x;scanf("%lf",&x);
        f[i].r=_f[n-i].r=x;
    }
    for (int i=1;i<=n;i++)g[i].r=(1.0/i/i);
    m=2*n;for (n=1;n<=m;n<<=1,L++);if(L)L--;
    for (int i=0;i<n;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<L);
    FFT(f,1),FFT(_f,1),FFT(g,1);
    for (int i=0;i<n;i++)e1[i]=f[i]*g[i],e2[i]=_f[i]*g[i];
    FFT(e1,-1),FFT(e2,-1);
    for (int i=0;i<=m/2;i++)
        printf("%.7lf\n",e1[i].r-e2[m/2-i].r);
}