BZOJ3527:[ZJOI2014]力(FFT)

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

Solution

不会markdown/LaTeX的下场就是只能像我一样在word里编辑好然后粘图

 

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define N (400000+100)
using namespace std;

double pi=acos(-1.0);
double q[N],g[N],ans[N];
int n,fn,l,r[N];
struct complex
{
    double x,y;
    complex(double xx=0,double yy=0)
    {
        x=xx; y=yy;
    }
}a[N],b[N];

complex operator + (complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
complex operator - (complex a,complex b){return complex(a.x-b.x,a.y-b.y);}
complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
complex operator / (complex a,double b){return complex(a.x/b,a.y/b);}

void FFT(int n,complex *a,int opt)
{
    for (int i=0; i<n; ++i)
        if (i<r[i])
            swap(a[i],a[r[i]]);
    for (int k=1; k<n; k<<=1)
    {
        complex wn=complex(cos(pi/k),opt*sin(pi/k));
        for (int i=0; i<n; i+=(k<<1))
        {
            complex w=complex(1,0);
            for (int j=0; j<k; ++j,w=w*wn)
            {
                complex x=a[i+j], y=w*a[i+j+k];
                a[i+j]=x+y; a[i+j+k]=x-y;
            }
        }
    }
    if (opt==-1) for (int i=0; i<n; ++i) a[i]=a[i]/n;
}

void Calc()
{
    memset(a,0,sizeof(a)); memset(b,0,sizeof(b));
    for (int i=0; i<=n; ++i) a[i].x=q[i];
    for (int i=0; i<=n; ++i) b[i].x=g[i];
    FFT(fn,a,1); FFT(fn,b,1);
    for (int i=0; i<=fn; ++i)
        a[i]=a[i]*b[i];
    FFT(fn,a,-1);
}

int main()
{
    scanf("%d",&n);
    for (int i=1; i<=n; ++i)
        scanf("%lf",&q[i]);
    for (int i=1; i<=n; ++i)
        g[i]=1.0/i/i;
    fn=1;
    while (fn<=n+n) fn<<=1, l++;
    for (int i=0; i<fn; ++i) 
        r[i]=(r[i>>1]>>1) | ((i&1)<<(l-1));
        
    Calc();
    for (int i=1; i<=n; ++i)
        ans[i]=a[i].x;
    for (int i=1; i<=n/2; ++i)
        swap(q[i],q[n-i+1]);
    Calc();
    for (int i=1; i<=n; ++i)
        printf("%.6lf\n",ans[i]-a[n-i+1].x);
}