bzoj千题计划167：bzoj3527: [Zjoi2014]力

Posted 2020-10-19 日拱一卒 功不唐捐

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

 

给出n个数qi，给出Fj的定义如下：

令Ei=Fi/qi，求Ei.

 

 

 

 以n=4为例：

 

 设数组a[],b[]

 

 令c[]=a[]反转

y[]=c[]*b[]

那么E[i]=x[i]-y[n-i-1]

 

#include<cmath>
#include<cstdio>
#include<algorithm>
 
using namespace std;
 
#define N 100001
const int M=1<<18;
 
const double pi=acos(-1);
 
struct Complex
{
    double x,y;
    Complex() { }
    Complex(double _x,double _y) : x(_x),y(_y) { }
    Complex operator + (Complex  & P)
    {
        Complex C;
        C.x=x+P.x;
        C.y=y+P.y;
        return C;
    }
    Complex operator - (Complex & P)
    {
        Complex C;
        C.x=x-P.x;
        C.y=y-P.y;
        return C;
    }
    Complex operator * (Complex & P)
    {
        Complex C;
        C.x=x*P.x-y*P.y;
        C.y=x*P.y+y*P.x;
        return C;
    }
};
typedef Complex E;
 
int n;
 
int r[M];
 
E b[M];
E x[M],y[M];
 
void fft(E *a,int f)
{
    for(int i=0;i<n;++i)
        if(i<r[i]) swap(a[i],a[r[i]]);
    for(int i=1;i<n;i<<=1)
    {
        E wn(cos(pi/i),f*sin(pi/i));
        for(int p=i<<1,j=0;j<n;j+=p)
        {
            E w(1,0);
            for(int k=0;k<i;++k,w=w*wn)
            {
                E x=a[j+k],y=w*a[j+k+i];
                a[j+k]=x+y; a[j+k+i]=x-y;
            }
        }
    }
}
 
int main()
{
    int m,t;
    scanf("%d",&n);
    n--; m=t=n;
    double p;
    for(int i=0;i<=n;++i) 
    {
        scanf("%lf",&p);
        x[i].x=p;
        y[n-i].x=p;
    }
    for(int i=1;i<=t;++i) b[i].x=1.0/i/i;
    m+=n;
    int bit=0;
    for(n=1;n<=m;n<<=1) bit++;
    for(int i=0;i<n;++i) r[i]=(r[i>>1]>>1)|((i&1)<<bit-1); 
    fft(b,1);
    fft(x,1);
    for(int i=0;i<n;++i) x[i]=x[i]*b[i];
    fft(x,-1);
    fft(y,1);
    for(int i=0;i<n;++i) y[i]=y[i]*b[i];
    fft(y,-1);
    for(int i=0;i<=t;++i) printf("%.3lf\\n",(x[i].x-y[t-i].x)/n);
}