【题目】求集合元素问题（1,2x+1,3X+1类)

某集合Ａ中的元素有以下特征：

(1)数1是A中的元素

(2)如果X是A中的元素,则2x+1,3x+1也是A中的元素

(3)除了条件(1),(2)以外的所有元素均不是A中的元素

[参考程序1]

uses crt,dos;

var a:array[1..10000]of longint;

    b:array[1..10000]of boolean;

    times,n,m,long,i:longint;

    hour1,minute1,second1,sec1001:word;

   hour2,minute2,second2,sec1002:word;

begin

     write('N=');readln(n);

{     gettime(hour1,minute1,second1,sec1001);

     times:=minute1*60+second1;

     writeln(minute1,':',second1);}

     fillchar(b,sizeof(b),0);

     a[1]:=1;m:=2;long:=1;

     while long<=n do begin

           for i:=1 to long do

               if (a[i]*2=m-1) or (a[i]*3=m-1) then

                 if not b[m] then begin

                  inc(long);a[long]:=m;b[m]:=true;break;

               end;

           inc(m);

     end;

{     gettime(hour2,minute2,second2,sec1002);

     times:=minute2*60+second2-times;

     writeln(minute2,':',second2);

     writeln('Ok! Uses Time: ',times);}

     for i:=1 to n do write(a[i],' ');

readln;

end.

[参考程序2]

uses crt;

const n=10000;

var a:array[1..n] of longint;

    i,j,k,l,y:longint;

begin

     clrscr;

     fillchar(a,sizeof(a),0);

     i:=1;j:=1;

     a[i]:=1;

     repeat

           y:=2*a[i]+1;

           k:=j;

           while y〈a[k] do begin

                 a[k+1]:=a[k];

                 k:=k-1;

           end;

           if y>a[k] then begin

              a[k+1]:=y;j:=j+1;

              end

           else for l:=k+1 to j do a[l]:=a[l+1];

           j:=j+1;

           a[j]:=3*a[i]+1;

           inc(i);

     until k>=n;

     for i:=1 to n do begin

         write(a[i],' ');

         if (i mod 10 =0 ) or (i=n) then writeln

     end;

end.

[参考程序3]

uses crt;

var a:array[1..10000]of longint;

    n,i,one,another,long,s,x,y:longint;

begin

     write('n=');readln(n);

     a[1]:=1;long:=1;one:=1;another:=1;

     while longy then begin s:=y;inc(another);end

                   else begin s:=x;inc(one);inc(another);end;

           inc(long);a[long]:=s;

     end;

     for i:=1 to n do write(a[i],' ');

end.

[参考程序4]

var n:integer;

    top,x:longint;

function init(x:longint):boolean;

begin

     if x=1 then init:=true

        else if((x-1)mod 2=0)and(init((x-1)div 2))

                     or((x-1)mod 3=0)and(init((x-1)div 3))then

             init:=true

             else init:=false;

end;

begin

     write('input n:');

     readln(n);

     x:=0;

     top:=0;

     while top< n do begin

           x:=x+1;

           if init(x) then

              top:=top+1;

              write(x:8);

           end;

     write('output end.');

     readln

end.