【UOJ34】高精度乘法（FFT）

题意：

思路：FFT模板，自带10倍常数

type cp=record
         x,y:double;
        end;
     arr=array[0..510000]of cp;
var a,b,cur:arr;
    n,m,n1,n2,i,j:longint;
    x:double;

function jia(a,b:cp;f:longint):cp;
begin
 if f=-1 then
 begin
  b.x:=-b.x; b.y:=-b.y;
 end;
 jia.x:=a.x+b.x;
 jia.y:=a.y+b.y;
end;

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

procedure swap(var x,y:cp);
var t:cp;
begin
 t:=x; x:=y; y:=t;
end;

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

procedure fft(var a:arr;n,f:longint);
var i,j,k,m:longint;
    w,u,v:cp;

begin
 i:=n>>1; j:=1;
 while j<n do
 begin
  if i<j then swap(a[i],a[j]);
  k:=n>>1;
  while k and i>0 do
  begin
   i:=i xor k;
   k:=k>>1;
  end;
  i:=i xor k;
  inc(j);
 end;
 m:=2;
 while m<=n do
 begin
  w.x:=cos(2*pi*f/m); w.y:=sin(2*pi*f/m);
  cur[0].x:=1; cur[0].y:=0;
  for i:=1 to m-1 do cur[i]:=mult(cur[i-1],w);
  i:=0;
  while i<n do
  begin
   j:=i;
   while j<i+(m>>1) do
   begin
    u:=a[j]; v:=mult(a[j+(m>>1)],cur[j-i]);
    a[j]:=jia(u,v,1);
    a[j+(m>>1)]:=jia(u,v,-1);
    inc(j);
   end;
   i:=i+m;
  end;
  m:=m<<1;
 end;
end;

begin
 assign(input,'uoj34.in'); reset(input);
 assign(output,'uoj34.out'); rewrite(output);
 readln(n1,n2);
 inc(n1); inc(n2);
 for i:=0 to n1-1 do
 begin
  read(x); a[i].x:=x;
 end;
 for i:=0 to n2-1 do
 begin
  read(x); b[i].x:=x;
 end;
 n:=max(n1,n2);
 m:=1;
 while m<=(n<<1) do m:=m<<1;
 fft(a,m,1); fft(b,m,1);
 for i:=0 to m do a[i]:=mult(a[i],b[i]);
 fft(a,m,-1);
 for i:=0 to n1+n2-2 do write(round(a[i].x/m),' ');
 close(input);
 close(output);
end.