【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=;
 var h1,h2,mi,c,a,head,vet,next:array[..]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-])*mi[y-x+] mod mo) mod mo;
  hash1:=(hash1+mo) mod mo;
 end;

 function hash2(x,y:longint):longint;
 begin
  hash2:=(h2[y]-int64(h2[x-])*mi[y-x+] 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<> 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:= to n do read(a[i]);
  mi[]:=;
  for i:= to n do mi[i]:=int64(mi[i-])* mod mo;
  for i:= to n do h1[i]:=(int64(h1[i-])*+a[i]) mod mo;
  for i:= to n div  do swap(a[i],a[n-i+]);
  for i:= to n do h2[i]:=(int64(h2[i-])*+a[i]) mod mo;
  for i:= to n do
  begin
   p:=; tot:=;
   for j:= to n div i do
   begin
    st:=i*(j-)+; ed:=i*j;
    t1:=hash1(st,ed); t2:=hash2(n-ed+,n-st+);
    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:=; c[]:=i; ans:=p;
   end;
   for j:= to n div i do
   begin
    st:=i*(j-)+; ed:=i*j;
    t1:=hash1(st,ed); t2:=hash2(n-ed+,n-st+);
    t3:=int64(t1)*t2 mod mo;
    inc(t3);
    head[t3]:=;
   end;
  end;
  writeln(ans,' ',l);
  for i:= to l- do write(c[i],' ');
  write(c[l]);
  close(input);
  close(output);
 end.