bzoj3527: [Zjoi2014]力 fft

题意:求\(E_i=\sum_{j=1}^{i-1}qj/{(i-j)^2}-\sum_{j=i+1}^{n}qj/{(i-j)^2}\)
题解:构造前几个Ei,可以发现\(E_i=a_i*b_{j-i}\),\(a_i=q_i\),\(b=-1/(n-1)^2-...-1/1^2-0+1/1^2+...+1/n^2\)
卷积搞一搞就行了

/**************************************************************
    Problem: 3527
    User: walfy
    Language: C++
    Result: Accepted
    Time:2748 ms
    Memory:29420 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
 
using namespace std;
 
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;
 
struct cd{
    db x,y;
    cd(db _x=0.0,db _y=0.0):x(_x),y(_y){}
    cd operator +(const cd &b)const{
        return cd(x+b.x,y+b.y);
    }
    cd operator -(const cd &b)const{
        return cd(x-b.x,y-b.y);
    }
    cd operator *(const cd &b)const{
        return cd(x*b.x - y*b.y,x*b.y + y*b.x);
    }
    cd operator /(const db &b)const{
        return cd(x/b,y/b);
    }
}a[N<<3],b[N<<3];
int rev[N<<3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void fft(cd *a,int n,int dft)
{
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        cd wn(cos(dft*pi/step),sin(dft*pi/step));
        for(int j=0;j<n;j+=step<<1)
        {
            cd wnk(1,0);
            for(int k=j;k<j+step;k++)
            {
                cd x=a[k];
                cd y=wnk*a[k+step];
                a[k]=x+y;a[k+step]=x-y;
                wnk=wnk*wn;
            }
        }
    }
    if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;
}
int main()
{
    int n;scanf("%d",&n);
    int sz=0;
    while((1<<sz)<n)sz++;
    sz++;getrev(sz);
    for(int i=0;i<=(1<<sz);i++)a[i]=b[i]=0;
    for(int i=0;i<n;i++)
    {
        double x;scanf("%lf",&x);
        a[i]=x;
    }
    int now=0;
    for(int i=n-1;i>=1;i--)b[now++]=-1.0/i/i;
    now++;
    for(int i=1;i<=n-1;i++)b[now++]=1.0/i/i;
    fft(a,(1<<sz),1);fft(b,(1<<sz),1);
    for(int i=0;i<=(1<<sz);i++)a[i]=a[i]*b[i];
    fft(a,(1<<sz),-1);
    for(int i=n-1;i<2*n-1;i++)printf("%lf\n",a[i].x);
    return 0;
}
/********************
 
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