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 
4
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：

 1 #!/usr/bin/perl
  2 my $debug=0; #该问题略微复杂，所以使用调试技术：如果$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。
  3 my $infile='C-large-practice.in';
  4 my $outfile='C-large.out';
  5 open $in,'<',$infile
  6         or die "Cannot open $infile:$!\n";
  7 open $out,'>',$outfile
  8         or die "Cannot open $outfile:$!\n";
  9 #建立哈希表
 10 my %dict=(
 11         2=>"abc",
 12         3=>"def",
 13         4=>"ghi",
 14         5=>"jkl",
 15         6=>"mno",
 16         7=>"pqrs",
 17         8=>"tuv",
 18         9=>"wxyz",
 19         0=>" "
 20 );
 21 
 22 if($debug){
 23         for(keys %dict){
 24                 print "$_=>'$dict{$_}'\n";
 25         }
 26 }
 27 chomp(my $N=<$in>);
 28 my $str;
 29 my $line;
 30 my $prev,$cur;
 31 $N=5 if $debug;
 32 for($i=1;$i<=$N;$i++){
 33         $str="";
 34         $prev="";
 35         chomp($line=<$in>);
 36         my $temp="";
 37         print "===================== For \$line='$line' =======================:\n" if $debug;
 38         print "length($line)=",length($line),"\n" if $debug;
 39         for(my $index=0;$index<length($line);$index++){
 40                 $cur=substr($line,$index,1);
 41                 $temp .=$cur if $debug;
 42                 print "\$prev='$prev',\$cur='$cur',\$index='$index',\$temp='$temp'\n" if $debug;
 43                 #if($cur eq $prev){ 
 44                 #       print "'$cur' eqs '$prev'\n" if $debug;
 45                 #       $str .=" ";
 46                 #}#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

 47                 LOOP1: while(($key,$value)=each %dict){
 48                         if((my $pos=index($value,"$cur"))!=-1){
 49                                 print "find '$cur' at $pos in '$value' for '$key'\n" if $debug;
 50                                 if( $key eq $prev){    #是否需要pause的关键不是当前字符是否和上一个字符相等，而是当前需要按下的按键和上一个按键是否相等
 51                                         print "'$key' eq '$prev' \n" if $debug;
 52                                         $str .=" ";
 53                                 }
 54                                 $prev=$key;#将当前按键存储到上一个按键中，以便下一次比较
 55                                 $str .= $key x ($pos+1);
 56                         #       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
 57                         }
 58                 }
 59         }
 60         if($debug){
 61                 print "Case #$i: $str\n";
 62         }
 63         else{
 64                 print $out "Case #$i: $str\n";
 65         }
 66 }

 

 

上传原题地址链接网站，结果正确。