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| author | Abigail <abigail@abigail.be> | 2021-05-30 15:53:23 +0200 |
|---|---|---|
| committer | Abigail <abigail@abigail.be> | 2021-05-30 15:53:23 +0200 |
| commit | 4a8f8b846b6a8b6cd25025cac07a43ffa8f6abe0 (patch) | |
| tree | 2b0f990b6e6a07194ee5daa33e3d1043164d95e8 | |
| parent | 627209bce4d1c1d616d3d24faa176733805110ae (diff) | |
| download | perlweeklychallenge-club-4a8f8b846b6a8b6cd25025cac07a43ffa8f6abe0.tar.gz perlweeklychallenge-club-4a8f8b846b6a8b6cd25025cac07a43ffa8f6abe0.tar.bz2 perlweeklychallenge-club-4a8f8b846b6a8b6cd25025cac07a43ffa8f6abe0.zip | |
Link to blog for week 114, part 1
| -rw-r--r-- | challenge-114/abigail/README.md | 1 | ||||
| -rw-r--r-- | challenge-114/abigail/blog.txt | 1 | ||||
| -rw-r--r-- | challenge-114/abigail/perl/ch-1.pl | 24 |
3 files changed, 4 insertions, 22 deletions
diff --git a/challenge-114/abigail/README.md b/challenge-114/abigail/README.md index ba6d2722ec..174d509623 100644 --- a/challenge-114/abigail/README.md +++ b/challenge-114/abigail/README.md @@ -22,6 +22,7 @@ Output: 1001 * [Perl](perl/ch-1.pl) ### Blog +[Next Palindrome Number](https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-114-1.html) ## [Higher Integer Set Bits](https://perlweeklychallenge.org/blog/perl-weekly-challenge-114/#TASK2) diff --git a/challenge-114/abigail/blog.txt b/challenge-114/abigail/blog.txt new file mode 100644 index 0000000000..087a442f76 --- /dev/null +++ b/challenge-114/abigail/blog.txt @@ -0,0 +1 @@ +https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-114-1.html diff --git a/challenge-114/abigail/perl/ch-1.pl b/challenge-114/abigail/perl/ch-1.pl index 4c1d752fed..510376da90 100644 --- a/challenge-114/abigail/perl/ch-1.pl +++ b/challenge-114/abigail/perl/ch-1.pl @@ -18,28 +18,8 @@ use experimental 'lexical_subs'; # # -# We are considering the following cases to determine the next -# palindromic integer: -# -# - If the number consists of just 9s, we add 2 to the number, resulting -# in 100001, where we have 1 less 0 than we had 9s. -# - Else, split the number into three parts: (NNNN)(M)(PPPP), where -# the first and third part are of equal lengths, and M is zero or one -# digit (zero digits of the input number is of even length, else the -# middle part is just one digit). -# -# Define nnnn as the reverse of NNNN, -# pppp as the reverse of PPPP, -# QQQQ as NNNN + 1, -# qqqq as the reverse of QQQQ, -# m = M + 1 (if M is one digit long) -# -# Now: -# - If nnnn > PPPP, then result is "NNNNMnnnn", else -# - If M is one digit long, then: -# - If M < 9, then the result is "NNNNmnnnn", else -# - the result is "QQQQ0qqqq", else -# - The result is "QQQQqqqq". +# See https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-114-1.html +# for a description of the algorithm. # while (<>) { |