[已解决] 如何编写四进制++的代码

substr函数

四进制++


my $number = 0;
如何编写$number++代码
1
2
3
10
11
12
13
20
21
22
23
30
31
32
33
100
101
102
103
110
111
112
113
120
121
...


my $number = 30;
如何编写$number++代码
31
32
33
100
101
102
103
110
111
112
113
120
121
...

jason680

回复 1# substr函数

Would you like this way with n4 function


$ cat n4.pl

use strict;
use warnings;

sub n4{
  local $_ = shift;
  while(m/[4-9]/){
    s/^([4-9])/0$1/;
    s|(\d)([4-9])|sprintf("%s",(($1?$1:0)+int($2/4)) . ($2%4))|ge;
  }
  return $_;
}

my $number = 0;
while($number <= 200){
  print "$number, ";
  $number = n4($number+1);
}
print "\n";

$ perl n4.pl
0, 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200,

   

substr函数

回复 2# jason680



Yes I did,I enjoyed it very much,thank you.
   

ghoull

1. #!/usr/bin/perl
   
2. 
   
3. use 5.010;
   
4. use strict;
   
5. use warnings;
   
6. 
   
7. my $format=4;
   
8. my $n=99;
   
9. for (my $var = 0; $var < $n; $var++)  {
   
10.         my $out='';
    
11.         while ($_>0) {
    
12.                 my $mo=$_%$format;
    
13.                 $_=($_-$mo)/$format;
    
14.                 $out=$mo . $out;
    
15.                 }
    
16.         say $out;
    
17. }

复制代码

substr函数

回复 4# ghoull


亲，谢谢回复!
哦！我试了试。不知道该怎么做比较好。怎么能每次 ++ 呢
my $number = 30;
如何$number++ => 31

substr函数

回复 6# walklan

亲，谢谢回复!