我们就建一颗线段树,线段树的每一个节点都是一颗平衡树,对于每个询问来说,我们就二分答案,

查询每个二分到的mid在这个区间里的rank,然后就行了

/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/ //By BLADEVIL
type
rec =record
left, right, root :longint;
end; var
n, m :longint;
a :array[..] of longint;
t :array[..] of rec;
bt :array[..] of longint;
b_size, b_left, b_right :array[..] of longint;
b_key :array[..] of longint;
tot :longint; function min(a,b:longint):longint;
begin
if a>b then min:=b else min:=a;
end; function max(a,b:longint):longint;
begin
if a>b then max:=a else max:=b;
end; procedure rotate_right(var t:longint);
var
k :longint;
begin
k:=b_left[t];
b_left[t]:=b_right[k];
b_right[k]:=t;
b_size[k]:=b_size[t];
b_size[t]:=b_size[b_right[t]]+b_size[b_left[t]]+;
t:=k;
end; procedure rotate_left(var t:longint);
var
k :longint;
begin
k:=b_right[t];
b_right[t]:=b_left[k];
b_left[k]:=t;
b_size[k]:=b_size[t];
b_size[t]:=b_size[b_right[t]]+b_size[b_left[t]]+;
t:=k;
end; procedure maintain(var t:longint; flag:boolean);
begin
if not flag then
begin
if b_size[b_left[b_left[t]]]>b_size[b_right[t]] then
rotate_right(b_left[t]) else
if b_size[b_right[b_left[t]]]>b_size[b_right[t]] then
begin
rotate_left(b_left[t]);
rotate_right(t);
end else exit;
end else
begin
if b_size[b_right[b_right[t]]]>b_size[b_left[t]] then
rotate_left(b_right[t]) else
if b_size[b_left[b_right[t]]]>b_size[b_left[t]] then
begin
rotate_right(b_right[t]);
rotate_left(t);
end else exit;
end;
maintain(b_left[t],false);
maintain(b_right[t],true);
maintain(t,true);
maintain(t,false);
end; procedure insert(var t:longint; v:longint);
begin
if t= then
begin
inc(tot);
t:=tot;
b_left[t]:=;
b_right[t]:=;
b_size[t]:=;
b_key[t]:=v;
end else
begin
inc(b_size[t]);
if v>=b_key[t] then
insert(b_right[t],v) else
insert(b_left[t],v);
maintain(t,v>=b_key[t]);
end;
end; function delete(var t:longint; v:longint):longint;
begin
dec(b_size[t]);
if (b_key[t]=v)
or (b_right[t]=) and (v>b_key[t])
or (b_left[t]=) and (v<b_key[t]) then
begin
delete:=b_key[t];
if (b_left[t]=) or (b_right[t]=) then
t:=b_right[t]+b_left[t] else
b_key[t]:=delete(b_left[t],v+);
end else
if v<b_key[t] then delete:=delete(b_left[t],v) else
delete:=delete(b_right[t],v);
end; function rank(var t:longint; v:longint):longint;
begin
if t= then exit();
if v<=b_key[t] then rank:=rank(b_left[t],v) else
rank:=rank(b_right[t],v)+b_size[b_left[t]]+;
end; procedure build(x,l,r:longint);
var
mid :longint;
i :longint;
begin
t[x].left:=l; t[x].right:=r;
t[x].root:=;
for i:=l to r do insert(t[x].root,a[i]);
if l=r then exit;
mid:=(l+r) div ;
build(x*,l,mid); build(x*+,mid+,r);
end; procedure change(x,y,z:longint);
var
mid :longint;
begin
mid:=delete(t[x].root,a[y]);
insert(t[x].root,z);
if t[x].left=t[x].right then exit;
with t[x] do mid:=(left+right) div ;
if y>mid then change(x*+,y,z) else change(x*,y,z);
end; function ask(x,l,r,k:longint):longint;
var
mid :longint;
begin
ask:=;
if (t[x].left=l) and (t[x].right=r) then
begin
ask:=rank(t[x].root,k);
exit;
end;
with t[x] do mid:=(left+right) div ;
if mid<l then ask:=ask(x*+,l,r,k) else
if mid>=r then ask:=ask(x*,l,r,k) else
ask:=ask(x*,l,mid,k)+ask(x*+,mid+,r,k);
end; function succ(var t:longint;v:longint):longint;
begin
if t= then exit(-);
if b_key[t]<v then succ:=succ(b_right[t],v) else
begin
succ:=succ(b_left[t],v);
if succ=- then succ:=b_key[t];
end;
end; function pred(var t:longint;v:longint):longint;
begin
if t= then exit(-);
if b_key[t]>v then pred:=pred(b_left[t],v) else
begin
pred:=pred(b_right[t],v);
if pred=- then pred:=b_key[t];
end;
end; function ask_succ(x,l,r,y:longint):longint;
var
mid :longint;
begin
if (t[x].left=l) and (t[x].right=r) then
begin
ask_succ:=succ(t[x].root,y);
if ask_succ=- then ask_succ:=maxlongint;
exit;
end;
with t[x] do mid:=(left+right) div ;
if l>mid then ask_succ:=ask_succ(x*+,l,r,y) else
if r<=mid then ask_succ:=ask_succ(x*,l,r,y) else
ask_succ:=min(ask_succ(x*,l,mid,y),ask_succ(x*+,mid+,r,y));
end; function ask_pred(x,l,r,y:longint):longint;
var
mid :longint;
begin
if (t[x].left=l) and (t[x].right=r) then
begin
ask_pred:=pred(t[x].root,y);
exit;
end;
with t[x] do mid:=(left+right) div ;
if l>mid then ask_pred:=ask_pred(x*+,l,r,y) else
if r<=mid then ask_pred:=ask_pred(x*,l,r,y) else
ask_pred:=max(ask_pred(x*,l,mid,y),ask_pred(x*+,mid+,r,y));
end; procedure query(x,y,k:longint);
var
l, r, mid :longint;
ans :longint;
xx :longint;
begin
l:=; r:=;
while l<=r do
begin
mid:=(l+r) div ;
xx:=ask(,x,y,mid);
if xx>=k- then
begin
ans:=mid;
r:=mid-;
end else l:=mid+;
end;
if ask(,x,y,ans)<>k- then
xx:=ask_pred(,x,y,ans-) else
xx:=ask_succ(,x,y,ans);
writeln(xx);
end; procedure init;
var
i :longint;
begin
read(n,m);
for i:= to n do read(a[i]);
readln;
build(,,n);
end; procedure main;
var
i :longint;
ch :char;
l, r, x :longint; begin
for i:= to m do
begin
read(ch);
if ch='Q' then
begin
readln(l,r,x);
query(l,r,x);
end else
begin
readln(l,x);
change(,l,x);
a[l]:=x;
end;
end;
end; begin
init;
main;
end.

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