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.

View Code