bzoj2821

其实和bzoj2724是一样的
都是先处理多个块的答案，然后多余部分暴力
空间要注意一下，还是O(nsqrt(n));

 var f:array[..,..] of longint;
     g:array[..,..] of longint;
     a,s,be:array[..] of longint;
     t,n,p,m,i,x,y,ans,size:longint;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 procedure pre;
   var i,j,k:longint;
   begin
     for i:= to t do
     begin
       k:=min(n,i*size);
       g[i]:=g[i-];
       for j:=(i-)*size+ to k do
         inc(g[i,a[j]]);
     end;
     for i:= to t do
     begin
       fillchar(s,sizeof(s),);
       for j:=(i-)*size+ to n do
       begin
         inc(s[a[j]]);
         if j=(be[j]-)*size+ then
           f[i,be[j]]:=f[i,be[j]-];
         if s[a[j]]>= then
           if s[a[j]] mod = then dec(f[i,be[j]]) else inc(f[i,be[j]]);
       end;
     end;
   end;

 procedure clear(x,y:longint);
   var i:longint;
   begin
     for i:=x to y do
       s[a[i]]:=;
   end;

 function ask(x,y:longint):longint;
   var i:longint;
   begin
     ask:=;
     if be[x]=be[y] then
     begin
       for i:=x to y do
       begin
         inc(s[a[i]]);
         if s[a[i]]>= then
           if s[a[i]] mod = then dec(ask) else inc(ask);
       end;
       clear(x,y);
     end
     else begin
       ask:=f[be[x]+,be[y]-];
       for i:=x to be[x]*size do
       begin
         if s[a[i]]= then s[a[i]]:=g[be[y]-,a[i]]-g[be[x],a[i]];
         inc(s[a[i]]);
         if s[a[i]]>= then
           if s[a[i]] mod = then dec(ask) else inc(ask);
       end;
       for i:=(be[y]-)*size+ to y do
       begin
         if s[a[i]]= then s[a[i]]:=g[be[y]-,a[i]]-g[be[x],a[i]];
         inc(s[a[i]]);
         if s[a[i]]>= then
           if s[a[i]] mod = then dec(ask) else inc(ask);
       end;
       clear(x,be[x]*size);
       clear((be[y]-)*size+,y);
     end;
   end;

 begin
   readln(n,p,m);
   size:=trunc(sqrt(n));
   for i:= to n do
   begin
     read(a[i]);
     be[i]:=(i-) div size+;
   end;
   t:=n div size;
   if n mod size<> then inc(t);
   pre;
   fillchar(s,sizeof(s),);
   for i:= to m do
   begin
     readln(x,y);
     x:=(x+ans) mod n+;
     y:=(y+ans) mod n+;
     if x>y then swap(x,y);
     ans:=ask(x,y);
     writeln(ans);
   end;
 end.