bzoj2743

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.