其实和bzoj1878类似
只不过要求的是区间内数量多于1个的数字种数
其实还是按照bzoj1878做
只不过我们是把每一种数字下一个出现的位置+1，并把这个位置置为0
var x,y,ans,p,last,a,c,next:array[..] of longint;
max,i,n,m,j:longint; function lowbit(x:longint):longint;
begin
exit(x and(-x));
end; function sum(x:longint):longint;
begin
sum:=;
while x<> do
begin
sum:=sum+c[x];
x:=x-lowbit(x);
end;
end; procedure add(x,t:longint);
begin
while x<=n do
begin
inc(c[x],t);
x:=x+lowbit(x);
end;
end; procedure swap(var a,b:longint);
var c:longint;
begin
c:=a;
a:=b;
b:=c;
end; procedure sort(l,r: longint);
var i,j,z: longint;
begin
i:=l;
j:=r;
z:=x[(l+r) div ];
repeat
while x[i]<z do inc(i);
while z<x[j] do dec(j);
if not(i>j) then
begin
swap(x[i],x[j]);
swap(y[i],y[j]);
swap(p[i],p[j]);
inc(i);
dec(j);
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end; begin
readln(n,max,m);
for i:= to n do
read(a[i]); for i:= to m do
begin
readln(x[i],y[i]);
p[i]:=i;
end;
sort(,m);
fillchar(last,sizeof(last),);
for i:=n downto do
begin
next[i]:=last[a[i]];
last[a[i]]:=i;
end;
for i:= to max do
if next[last[i]]<> then add(next[last[i]],); j:=;
for i:= to n do
begin
while x[j]=i do
begin
ans[p[j]]:=sum(y[j])-sum(x[j]-);
inc(j);
end;
if next[i]<> then add(next[i],-);
if next[next[i]]<> then add(next[next[i]],);
end;
for i:= to m do
writeln(ans[i]);
end.