Google APAC----Africa 2010, Qualification Round(Problem C. T9 Spelling)----Perl 解法

原题地址链接：https://code.google.com/codejam/contest/351101/dashboard#s=p2

问题描述：

Problem

The Latin alphabet contains 26 characters and telephones only have ten digits on the keypad. We would like to make it easier to write a message to your friend using a sequence of keypresses to indicate the desired characters. The letters are mapped onto the digits as shown below. To insert the character B for instance, the program would press22. In order to insert two characters in sequence from the same key, the user must pause before pressing the key a second time. The space character ' ' should be printed to indicate a pause. For example, 2 2 indicates AA whereas 22 indicates B.

Input

The first line of input gives the number of cases, N. N test cases follow. Each case is a line of text formatted as

   desired_message

Each message will consist of only lowercase characters a-z and space characters ' '. Pressing zero emits a space.

Output

For each test case, output one line containing "Case #x: " followed by the message translated into the sequence of keypresses.

Limits

1 ≤ N ≤ 100.

Small dataset

1 ≤ length of message in characters ≤ 15.

Large dataset

1 ≤ length of message in characters ≤ 1000.

Sample

Input 

hi
yes
foo  bar
hello world

Output
Case #1: 44 444
Case #2: 999337777
Case #3: 333666 6660 022 2777
Case #4: 4433555 555666096667775553

Perl算法：

  #!/usr/bin/perl
   my $debug=; #该问题略微复杂，所以使用调试技术：如果$debug为1，则结果显示在标准输出上，而不输出到 .out 文件内；如果$debug 为0，则结果直接输出到 .out 文件中。
   my $infile='C-large-practice.in';
   my $outfile='C-large.out';
   open $in,'<',$infile
           or die "Cannot open $infile:$!\n";
   open $out,'>',$outfile
           or die "Cannot open $outfile:$!\n";
   #建立哈希表
  my %dict=(
          =>"abc",
          =>"def",
          =>"ghi",
          =>"jkl",
          =>"mno",
          =>"pqrs",
          =>"tuv",
          =>"wxyz",
          =>" "
  );

  if($debug){
          for(keys %dict){
                  print "$_=>'$dict{$_}'\n";
          }
  }
  chomp(my $N=<$in>);
  my $str;
  my $line;
  my $prev,$cur;
  $N= if $debug;
  for($i=;$i<=$N;$i++){
          $str="";
          $prev="";
          chomp($line=<$in>);
          my $temp="";
          print "===================== For \$line='$line' =======================:\n" if $debug;
          print "length($line)=",length($line),"\n" if $debug;
          for(my $index=;$index<length($line);$index++){
                  $cur=substr($line,$index,);
                  $temp .=$cur if $debug;
                  print "\$prev='$prev',\$cur='$cur',\$index='$index',\$temp='$temp'\n" if $debug;
                  #if($cur eq $prev){
                  #       print "'$cur' eqs '$prev'\n" if $debug;
                  #       $str .=" ";
                  #}#是否需要pause的关键不是 当前字符 $cur 与上一个字符 $prev 是否相等，而是当前需要按下的按键 $key 是否与上一个按键 $prev 是否相等，因此注释掉第一次使用的算法

                  LOOP1: while(($key,$value)=each %dict){
                          if((my $pos=index($value,"$cur"))!=-){
                                  print "find '$cur' at $pos in '$value' for '$key'\n" if $debug;
                                  if( $key eq $prev){    #是否需要pause的关键不是当前字符是否和上一个字符相等，而是当前需要按下的按键和上一个按键是否相等
                                          print "'$key' eq '$prev' \n" if $debug;
                                          $str .=" ";
                                  }
                                  $prev=$key;#将当前按键存储到上一个按键中，以便下一次比较
                                  $str .= $key x ($pos+);
                          #       last LOOP1; #即使已经找到了匹配的键值，也不能提前跳出循环，http://www.cnblogs.com/dongling/p/5705224.html 这篇随笔解释了原因
                          }
                  }
          }
          if($debug){
                  print "Case #$i: $str\n";
          }
          else{
                  print $out "Case #$i: $str\n";
          }
  }

上传原题地址链接网站，结果正确。