3527: [Zjoi2014]力 - BZOJ

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=;
 type
     cp=record
         x,y:double;
     end;
     arr=array[..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= then exit;
     dft(a,s,t+);dft(a,s+<<t,t+);
     for i:= to n>>t>>- do
         begin
             p:=i<<t<<+s;
             wt:=w[i<<t]*a[p+<<t];
             tt[i]:=a[p]+wt;
             tt[i+n>>t>>]:=a[p]-wt;
         end;
     for i:= to n>>t- do
         a[s+i<<t]:=tt[i];
 end;

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

 begin
     main;
 end.