把质因子只有2,3,5的数称为Ugly数，求第k大的Ugly数。（1是第一个）

用堆保存数字，开始堆中只有1，每次删除最小值，把最小值的2倍，3倍，5倍插入到堆中，执行k次就得到结果。

注意要判重，用个hash就行，有点羡慕C++的map了。

1 program pku1338(input,output);   2 type   3    longint = int64;   4    node       = ^link;   5    link       = record   6         worth : longint;   7         next  : node;   8          end;       9 var  10    heap      : array[0..10000] of longint;  11    hash      : array[0..65537] of node;  12    n,tot  : longint;  13    answer : longint;  14 procedure swap(var aa,bb: longint );  15 var  16    tt : longint;  17 begin  18    tt:=aa;  19    aa:=bb;  20    bb:=tt;  21 end; {
      swap }  22 function find(x :longint ):boolean;  23 var  24    t : node;  25 begin  26    t:=hash[x mod 65537];  27    while t<>nil do  28    begin  29       if t^.worth=x then  30      exit(true);  31       t:=t^.next;  32    end;  33    exit(false);  34 end; {
      find }  35 procedure inn(x    :longint );  36 var  37    t : node;  38 begin  39    new(t);  40    t^.worth:=x;  41    t^.next:=hash[x mod 65537];  42    hash[x mod 65537]:=t;  43 end; {
      inn }  44 procedure up(x: longint );  45 begin  46    while (heap[x]
      
       >1])and(x<>1) do  47    begin  48       swap(heap[x],heap[x>>1]);  49       x:=x>>1;  50    end;  51 end; {
      up }  52 procedure down(x :longint );  53 begin  54    while (x*2<=tot)and((heap[x]>heap[x*2])or(heap[x]>heap[x*2+1])) do  55    begin  56       if heap[x<<1]>heap[x*2+1] then  57       begin  58      swap(heap[x],heap[x*2+1]);  59      x:=x*2+1;  60       end  61       else  62       begin  63      swap(heap[x],heap[x*2]);  64      x:=x<<1;  65       end;  66    end;  67 end; {
      down }  68 procedure insect(x :longint );  69 begin  70    inc(tot);  71    heap[tot]:=x;  72    up(tot);  73 end; {
      insect }  74 function delete():longint;  75 begin  76    delete:=heap[1];  77    swap(heap[1],heap[tot]);  78    heap[tot]:=maxlongint>>1;  79    dec(tot);  80    down(1);  81 end; {
      delete }  82 procedure main;  83 var  84    i : dword;  85 begin  86    fillchar(heap,sizeof(heap),63);  87    for i:=1 to 65537 do  88       hash[i]:=nil;  89    answer:=1;  90    tot:=1;  91    heap[1]:=1;  92    inn(1);  93    for i:=1 to n do  94    begin  95       answer:=delete();  96       if not find(answer*2) then  97       begin  98      inn(answer*2);  99      insect(answer*2); 100       end; 101       if not find(answer*3) then 102       begin 103      inn(answer*3); 104      insect(answer*3); 105       end; 106       if not find(answer*5) then 107       begin 108      inn(answer*5); 109      insect(answer*5); 110       end; 111    end; 112    writeln(answer); 113 end; {
      main } 114 begin 115    readln(n); 116    while n<>0 do 117    begin 118       main; 119       readln(n); 120    end; 121 end.