3527: [Zjoi2014]力

题目大意：给出n个数qi，定义 Fj为

      

    令 Ei=Fi/qi，求Ei。

看了很久题解，终于有些眉目，因为知道要用FFT，所以思路就很直了

其实我们就是要±1/(j-i)^2 ( j-i大于0时为正，小于0时为负 ) 和 qi 的乘积要算到j这个位置上，这个满足卷积，所以用FFT优化，但是j-i有负数，所以我们就加上一个n

于是设pi={

i>n,1/(i-n)^2

i<n,-1/(n-i)^2

其他,0

}

然后就套FFT模板就行了

const
    maxn=800100;
type
    cp=record
        x,y:double;
    end;
    arr=array[0..maxn]of cp;
var
    a,b,w,tt:arr;
    n,m:longint;

operator -(a,b:cp)c:cp;
begin
    c.x:=a.x-b.x;
    c.y:=a.y-b.y;
end;

operator +(a,b:cp)c:cp;
begin
    c.x:=a.x+b.x;
    c.y:=a.y+b.y;
end;

operator *(a,b:cp)c:cp;
begin
    c.x:=a.x*b.x-a.y*b.y;
    c.y:=a.x*b.y+a.y*b.x;
end;

procedure dft(var a:arr;s,t:longint);
var
    i,p:longint;
    wt:cp;
begin
    if n>>t=1 then exit;
    dft(a,s,t+1);dft(a,s+1<<t,t+1);
    for i:=0 to n>>t>>1-1 do
        begin
            p:=i<<t<<1+s;
            wt:=w[i<<t]*a[p+1<<t];
            tt[i]:=a[p]+wt;
            tt[i+n>>t>>1]:=a[p]-wt;
        end;
    for i:=0 to n>>t-1 do
        a[s+i<<t]:=tt[i];
end;

procedure main;
var
    i:longint;
begin
    read(m);
    for i:=0 to m-1 do read(a[i].x);
    for i:=1 to m-1 do b[i].x:=-1/(m-i)/(m-i);
    for i:=m+1 to m<<1-1 do b[i].x:=1/(i-m)/(i-m);
    n:=1;
    while n<m<<1 do n:=n<<1;
    for i:=0 to n-1 do w[i].x:=cos(pi*2*i/n);
    for i:=0 to n-1 do w[i].y:=sin(pi*2*i/n);
    dft(a,0,0);dft(b,0,0);
    for i:=0 to n-1 do a[i]:=a[i]*b[i];
    for i:=0 to n-1 do w[i].y:=-w[i].y;
    dft(a,0,0);
    for i:=m to m<<1-1 do writeln(a[i].x/n:0:3);
end;

begin
    main;
end.