poj1338

【题意】

回答第n个丑数是多少

丑数即为质因数只有2、3、5的数

【输入】

多次询问，询问以一个0结束

每次询问只有一个n

【输出】

回答每次询问

这个预处理一下前1500个丑数，回答的时候直接回答就好了

我挂了个堆来生成丑数，实际上没必要……

program poj1338;var  all,tot,n,i,j,k:longint;  ans:array [0..1501] of int64;  heap:array [0..10001] of int64;procedure swap (var a,b:int64)inline;var  i:int64;begin  i:=a;  a:=b;  b:=i;end;procedure insert (now:int64);inline;var  i:longint;begin  inc(tot);  heap[tot]:=now;  i:=tot;  while (i>1)and(heap[i]<heap[i div 2]) do    begin      swap(heap[i],heap[i div 2]);      i:=i div 2;    end;end;procedure delete;inline;var  i:longint;begin  if tot=1 then    begin      heap[1]:=heap[0];      dec(tot);      exit;    end;  swap(heap[1],heap[tot]);  heap[tot]:=heap[0];  dec(tot);  i:=1;  while (heap[i*2]<heap[i])or(heap[i*2+1]<heap[i]) do    if heap[i*2]<heap[i*2+1] then      begin        swap(heap[i],heap[i*2]);        i:=i*2;      end                             else      begin        swap(heap[i],heap[i*2+1]);        i:=i*2+1;      end;end;begin  fillchar(heap,sizeof(heap),63);  tot:=0;  ans[1]:=1;  insert(2);  insert(3);  insert(5);  for i:=2 to 1500 do    begin      while heap[1]=ans[i-1] do delete;      ans[i]:=heap[1];      delete;      insert(ans[i]*2);      insert(ans[i]*3);      insert(ans[i]*5);    end;  repeat    read(n);    if n=0 then break;    writeln(ans[n]);  until false;end.