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
1 const 2 maxn=800100; 3 type 4 cp=record 5 x,y:double; 6 end; 7 arr=array[0..maxn]of cp; 8 var 9 a,b,w,tt:arr; 10 n,m:longint; 11 12 operator -(a,b:cp)c:cp; 13 begin 14 c.x:=a.x-b.x; 15 c.y:=a.y-b.y; 16 end; 17 18 operator +(a,b:cp)c:cp; 19 begin 20 c.x:=a.x+b.x; 21 c.y:=a.y+b.y; 22 end; 23 24 operator *(a,b:cp)c:cp; 25 begin 26 c.x:=a.x*b.x-a.y*b.y; 27 c.y:=a.x*b.y+a.y*b.x; 28 end; 29 30 procedure dft(var a:arr;s,t:longint); 31 var 32 i,p:longint; 33 wt:cp; 34 begin 35 if n>>t=1 then exit; 36 dft(a,s,t+1);dft(a,s+1<<t,t+1); 37 for i:=0 to n>>t>>1-1 do 38 begin 39 p:=i<<t<<1+s; 40 wt:=w[i<<t]*a[p+1<<t]; 41 tt[i]:=a[p]+wt; 42 tt[i+n>>t>>1]:=a[p]-wt; 43 end; 44 for i:=0 to n>>t-1 do 45 a[s+i<<t]:=tt[i]; 46 end; 47 48 procedure main; 49 var 50 i:longint; 51 begin 52 read(m); 53 for i:=0 to m-1 do read(a[i].x); 54 for i:=1 to m-1 do b[i].x:=-1/(m-i)/(m-i); 55 for i:=m+1 to m<<1-1 do b[i].x:=1/(i-m)/(i-m); 56 n:=1; 57 while n<m<<1 do n:=n<<1; 58 for i:=0 to n-1 do w[i].x:=cos(pi*2*i/n); 59 for i:=0 to n-1 do w[i].y:=sin(pi*2*i/n); 60 dft(a,0,0);dft(b,0,0); 61 for i:=0 to n-1 do a[i]:=a[i]*b[i]; 62 for i:=0 to n-1 do w[i].y:=-w[i].y; 63 dft(a,0,0); 64 for i:=m to m<<1-1 do writeln(a[i].x/n:0:3); 65 end; 66 67 begin 68 main; 69 end.
