【BZOJ2081】Beads（哈希表）

题意：

翻转是指其中一段长度为k的子串全部翻转

n<=200000 a[i]<=n

思路：枚举k，直接哈希判充即可

时间复杂度是n/i求和，根据定理可得是O（n log n）级别的

单哈双哈都可能被卡，我用的是单哈+哈希表判重

const mo=250109;
var h1,h2,mi,c,a,head,vet,next:array[0..1000000]of longint;
    n,i,p,j,st,ed,t1,t2,t3,l,ans,tot:longint;

function hash1(x,y:longint):longint;
begin
 hash1:=(h1[y]-int64(h1[x-1])*mi[y-x+1] mod mo) mod mo;
 hash1:=(hash1+mo) mod mo;
end;

function hash2(x,y:longint):longint;
begin
 hash2:=(h2[y]-int64(h2[x-1])*mi[y-x+1] mod mo) mod mo;
 hash2:=(hash2+mo) mod mo;
end;

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

procedure add(a,b:longint);
begin
 inc(tot);
 next[tot]:=head[a];
 vet[tot]:=b;
 head[a]:=tot;
end;

function judge(u,x:longint):boolean;
var e,v:longint;
begin
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if v=x then exit(false);
  e:=next[e];
 end;
 exit(true);
end;

begin
 assign(input,'bzoj2081.in'); reset(input);
 assign(output,'bzoj2081.out'); rewrite(output);
 readln(n);
 for i:=1 to n do read(a[i]);
 mi[0]:=1;
 for i:=1 to n do mi[i]:=int64(mi[i-1])*200291 mod mo;
 for i:=1 to n do h1[i]:=(int64(h1[i-1])*200291+a[i]) mod mo;
 for i:=1 to n div 2 do swap(a[i],a[n-i+1]);
 for i:=1 to n do h2[i]:=(int64(h2[i-1])*200291+a[i]) mod mo;
 for i:=1 to n do
 begin
  p:=0; tot:=0;
  for j:=1 to n div i do
  begin
   st:=i*(j-1)+1; ed:=i*j;
   t1:=hash1(st,ed); t2:=hash2(n-ed+1,n-st+1);
   t3:=int64(t1)*t2 mod mo;
   inc(t3); inc(t1); inc(t2);
   if judge(t3,t1) then
   begin
    inc(p);
    add(t3,t1);
    add(t3,t2);
   end;
  end;
  if p=ans then begin inc(l); c[l]:=i; end;
  if p>ans then
  begin
   l:=1; c[1]:=i; ans:=p;
  end;
  for j:=1 to n div i do
  begin
   st:=i*(j-1)+1; ed:=i*j;
   t1:=hash1(st,ed); t2:=hash2(n-ed+1,n-st+1);
   t3:=int64(t1)*t2 mod mo;
   inc(t3);
   head[t3]:=0;
  end;
 end;
 writeln(ans,' ',l);
 for i:=1 to l-1 do write(c[i],' ');
 write(c[l]);
 close(input);
 close(output);
end.