我们可以贪心的分析，每个点的入度如果是0，那么这个点不可能
被用来更新答案，那么我们每次找入度为0的点，将他去掉，如果他连的
点没有被更新过答案，那么更新答案，去掉该点，环的时候最后处理就行了
/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/
//By BLADEVIL
var
n, ans, cnt :longint;
other, sum :array[..] of longint;
flag :array[..] of boolean;
que :array[..] of longint;
procedure init;
var
i :longint;
begin
readln(n);
for i:= to n do
begin
read(other[i]);
inc(sum[other[i]]);
end;
end;
procedure main;
var
h, t, cur :longint;
i, j :longint;
begin
t:=;
for i:= to n do
if sum[i]= then
begin
inc(t);
que[t]:=i;
end;
h:=;
while h<=t do
begin
cur:=que[h];
inc(h);
if (not flag[cur]) and (not flag[other[cur]]) then
begin
inc(ans);
flag[other[cur]]:=true;
dec(sum[other[other[cur]]]);
if sum[other[other[cur]]]= then
begin
inc(t);
que[t]:=other[other[cur]];
end;
end;
flag[cur]:=true;
end;
for i:= to n do
if not flag[i] then
begin
cnt:=;
flag[i]:=true;
j:=i;
while other[j]<>i do
begin
flag[other[j]]:=true;
inc(cnt);
j:=other[j];
end;
inc(ans,cnt div );
end;
writeln(ans);
end;
begin
init;
main;
end.