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-rw-r--r--challenge-113/abigail/awk/ch-1.awk16
-rw-r--r--challenge-113/abigail/bash/ch-1.sh31
-rw-r--r--challenge-113/abigail/c/ch-1.c16
-rw-r--r--challenge-113/abigail/lua/ch-1.lua30
-rw-r--r--challenge-113/abigail/node/ch-1.js16
-rw-r--r--challenge-113/abigail/perl/ch-1.pl65
-rw-r--r--challenge-113/abigail/python/ch-1.py17
-rw-r--r--challenge-113/abigail/ruby/ch-1.rb18
8 files changed, 111 insertions, 98 deletions
diff --git a/challenge-113/abigail/awk/ch-1.awk b/challenge-113/abigail/awk/ch-1.awk
index 91923aadb6..950a1052d0 100644
--- a/challenge-113/abigail/awk/ch-1.awk
+++ b/challenge-113/abigail/awk/ch-1.awk
@@ -8,19 +8,27 @@
# Run as: awk -f ch-1.awk < input-file
#
+#
+# For a description of the algorithm, and the proof why this is correct:
+# https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+#
+
BEGIN {
- split ("0 1 2 1 0 2 6 3 8", tens)
+ split ("1 2 1 2 5 2 1 2 1", gcds)
}
{
N = $1
D = $2
- D10 = D == 0 ? 100 : D * 10
- if (N >= D10 || N % (D == 0 ? 10 : D) == 0) {
+ if (D == 0) {
+ print (N >= 100 || N % 10 == 0 ? 1 : 0)
+ next
+ }
+ if (N >= 10 * D) {
print 1
next
}
- for (i = 1; i <= tens [D]; i ++) {
+ for (i = 0; i < D / gcds [D]; i ++) {
T = N - 10 * i - D
if (T >= 0 && T % D == 0) {
print 1
diff --git a/challenge-113/abigail/bash/ch-1.sh b/challenge-113/abigail/bash/ch-1.sh
index 7183523485..d3e60eddf4 100644
--- a/challenge-113/abigail/bash/ch-1.sh
+++ b/challenge-113/abigail/bash/ch-1.sh
@@ -8,22 +8,31 @@
# Run as: bash ch-1.sh < input-file
#
-tens=(0 0 1 2 1 0 2 6 3 8)
+#
+# For a description of the algorithm, and the proof why this is correct:
+# https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+#
+
+gcds=(0 1 2 1 2 5 2 1 2 1)
while read N D
-do ((D10 = D == 0 ? 100 : 10 * D))
- if ((N >= D10 || (N % (D == 0 ? 10 : D) == 0)))
+do if ((D == 0))
+ then if ((N >= 100 || N % 10 == 0))
+ then echo 1
+ else echo 0
+ fi
+ continue
+ fi
+ if ((N >= D * 10))
then echo 1
continue
fi
- for ((i = 1; i <= ${tens[$D]}; i ++))
- do ((T = N - 10 * i - D))
- if ((T >= 0 && T % D == 0))
- then echo 1
- continue 2
- fi
+ for ((i = 0; i < D / gcds[D]; i ++))
+ do ((T = N - 10 * i - D))
+ if ((T >= 0 && T % D == 0))
+ then echo 1
+ continue 2
+ fi
done
echo 0
done
-
-
diff --git a/challenge-113/abigail/c/ch-1.c b/challenge-113/abigail/c/ch-1.c
index e673fa7281..dac9106ef7 100644
--- a/challenge-113/abigail/c/ch-1.c
+++ b/challenge-113/abigail/c/ch-1.c
@@ -11,23 +11,31 @@
* Run as: cc -o ch-1.o ch-1.c; ./ch-1.o < input-file
*/
+/*
+ * For a description of the algorithm, and the proof why this is correct:
+ * https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+ */
+
typedef long long number;
typedef short digit;
-unsigned short tens [] = {0, 0, 1, 2, 1, 0, 2, 6, 3, 8};
+unsigned short gcds [] = {0, 1, 2, 1, 2, 5, 2, 1, 2, 1};
int main (void) {
number N;
digit D;
while (scanf ("%lld %hd", &N, &D) == 2) {
- digit D10 = D == 0 ? 100 : 10 * D;
- if (N >= D10 || (N % (D ? D : 10) == 0)) {
+ if (D == 0) {
+ printf ("%d\n", N >= 100 || N % 10 == 0 ? 1 : 0);
+ continue;
+ }
+ if (N >= D * 10) {
printf ("1\n");
continue;
}
bool valid = false;
- for (unsigned short i = 1; i <= tens [D]; i ++) {
+ for (unsigned short i = 0; i < D / gcds [D]; i ++) {
number T = N - 10 * i - D;
if (T >= 0 && T % D == 0) {
printf ("1\n");
diff --git a/challenge-113/abigail/lua/ch-1.lua b/challenge-113/abigail/lua/ch-1.lua
index 36bd8e8afd..4de70d106c 100644
--- a/challenge-113/abigail/lua/ch-1.lua
+++ b/challenge-113/abigail/lua/ch-1.lua
@@ -8,29 +8,35 @@
-- Run as: lua ch-1.lua < input-file
--
-local tens = {0, 1, 2, 1, 0, 2, 6, 3, 8}
+--
+-- For a description of the algorithm, and the proof why this is correct:
+-- https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+--
+
+local gcds = {1, 2, 1, 2, 5, 2, 1, 2, 1}
for line in io . lines () do
local _, _, N, D = line : find ("([0-9]+)%s+([0-9])")
N = tonumber (N)
D = tonumber (D)
- local D10 = 10 * D
if D == 0
- then D10 = 100
+ then if N >= 100 or N % 10 == 0
+ then print (1)
+ else print (0)
+ end
+ goto end_loop
end
- if (N >= D10) or (D == 0 and N % 10 == 0)
- or (D > 0 and N % D == 0)
+
+ if N >= D * 10
then print (1)
goto end_loop
end
- if D > 0
- then for i = 1, tens [D]
- do local T = N - 10 * i - D
- if T >= 0 and T % D == 0
- then print (1)
- goto end_loop
- end
+ for i = 0, D / gcds [D] - 1
+ do local T = N - 10 * i - D
+ if T >= 0 and T % D == 0
+ then print (1)
+ goto end_loop
end
end
diff --git a/challenge-113/abigail/node/ch-1.js b/challenge-113/abigail/node/ch-1.js
index 2ca4f6a0e1..e755524470 100644
--- a/challenge-113/abigail/node/ch-1.js
+++ b/challenge-113/abigail/node/ch-1.js
@@ -8,18 +8,26 @@
// Run as: node ch-1.js < input-file
//
-let tens = [0, 0, 1, 2, 1, 0, 2, 6, 3, 8];
+//
+// For a description of the algorithm, and the proof why this is correct:
+// https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+//
+
+let gcds = [0, 1, 2, 1, 2, 5, 2, 1, 2, 1];
require ('readline')
. createInterface ({input: process . stdin})
. on ('line', _ => {
let [N, D] = _ . split (/\s+/) . map (_ => +_)
- let D10 = D == 0 ? 100 : 10 * D
- if (N >= D10 || (N % (D == 0 ? 10 : D) == 0)) {
+ if (D == 0) {
+ console . log (N >= 100 || N % 10 == 0 ? 1 : 0)
+ return
+ }
+ if (N >= D * 10) {
console . log (1)
return
}
- for (let i = 1; i <= tens [D]; i ++) {
+ for (let i = 0; i < D / gcds [D]; i ++) {
let T = N - 10 * i - D
if (T >= 0 && T % D == 0) {
console . log (1)
diff --git a/challenge-113/abigail/perl/ch-1.pl b/challenge-113/abigail/perl/ch-1.pl
index bb45261f33..78a0eb0947 100644
--- a/challenge-113/abigail/perl/ch-1.pl
+++ b/challenge-113/abigail/perl/ch-1.pl
@@ -18,68 +18,23 @@ use experimental 'lexical_subs';
#
#
-# For any of the digits $D, more numbers $N can be written as a sum
-# of positive integers, each containing the digit $D at least once.
+# For a description of the algorithm, and the proof why this is correct:
+# https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
#
-# Let $D10 = $D == 0 ? 100 : $D * 10.
-#
-# It should be obvious that if $D > 0 and $N >= $D10, we can write $N as a
-# sum of integers each containing at least one $D:
-#
-# For $D > 0:
-# Let $N = $D * k + i, k >= 10 and 0 <= i < $D,
-# then $N = $D10 + i + (k - 10) * $D.
-#
-# $D10 + i is a number containing exactly one $D, so $N can be
-# written as a sum of (k - 9) integers: 10 * $D + 1 and k - 10 $D's.
-#
-# For $D = 0:
-# Let $N = 100 + i + k * 10, for some 0 <= i < 10.
-#
-# 100 + i will contain a 0, so $N can be written as a sum of
-# integers each containing a 0.
-#
-# It should also be clear that for $D > 0, if $N % $D == 0, $N can be
-# written as a sum of integers containing $D, as $N is then a multiple of $D.
-#
-# For $N < $D10, $N can be written as a sum of positive integers
-# containing at least one $D, iff:
-#
-# 1) $N is divisible by $D (or $D10 if $D == 0) OR
-# 2) $N - 10 * l is non-negative and divisible by $D,
-# for some 0 < l < $D.
-#
-# In case 1), $N is the sum of $N/$D $Ds.
-# In case 2), $N is the sum of 10 * l + $D, and ($N - 10 * l - $D)/$D $Ds.
-#
-#
-# We can further restrict which ls we have to check for a given $D:
-#
-# $D | l
-# -----+-----
-# 0 |
-# 1 |
-# 2 | 1
-# 3 | 1, 2
-# 4 | 1
-# 5 |
-# 6 | 1, 2
-# 7 | 1, 2, 3, 4, 5, 6
-# 8 | 1, 2, 3
-# 9 | 1, 2, 3, 4, 5, 6, 7, 8
-#
-# These are all the ls < $D so that $D does not divide 10 * l.
-my @tens = (0, 0, 1, 2, 1, 0, 2, 6, 3, 8);
+my @gcds = (0, 1, 2, 1, 2, 5, 2, 1, 2, 1);
MAIN: while (<>) {
- my ($N, $D) = /[0-9]+/g;
- my $D10 = $D == 0 ? 100 : 10 * $D;
- if ($N >= $D10 || ($N % ($D || 10) == 0)) {
+ my ($N, $D) = split;
+ if ($D == 0) {
+ say $N >= 100 || $N % 10 == 0 ? 1 : 0;
+ next MAIN;
+ }
+ if ($N >= $D * 10) {
say 1;
next MAIN;
}
- for (my $i = 1; $i <= $tens [$D]; $i ++) {
+ for (my $i = 0; $i < $D / $gcds [$D]; $i ++) {
my $T = $N - 10 * $i - $D;
if ($T >= 0 && $T % $D == 0) {
say 1;
diff --git a/challenge-113/abigail/python/ch-1.py b/challenge-113/abigail/python/ch-1.py
index 7bf1b0db11..ebb8821ddc 100644
--- a/challenge-113/abigail/python/ch-1.py
+++ b/challenge-113/abigail/python/ch-1.py
@@ -8,20 +8,29 @@
# Run as: python ch-1.py < input-file
#
+#
+# For a description of the algorithm, and the proof why this is correct:
+# https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+#
+
import fileinput
-tens = [0, 0, 1, 2, 1, 0, 2, 6, 3, 8]
+gcds = [0, 1, 2, 1, 2, 5, 2, 1, 2, 1]
for line in fileinput . input ():
(N, D) = line . split ();
N = int (N)
D = int (D)
- D10 = 100 if D == 0 else 10 * D
- if N >= D10 or (D == 0 and N % 10 == 0) or (D > 0 and N % D == 0):
+ if D == 0:
+ print (1 if N >= 100 or N % 10 == 0 else 0)
+ continue
+
+ if N >= D * 10:
print (1)
continue
+
done = False
- for i in range (1, tens [D] + 1):
+ for i in range (0, D // gcds [D]):
T = N - 10 * i - D
if T >= 0 and T % D == 0:
print (1)
diff --git a/challenge-113/abigail/ruby/ch-1.rb b/challenge-113/abigail/ruby/ch-1.rb
index 4f44bf5130..dadd1f1a9c 100644
--- a/challenge-113/abigail/ruby/ch-1.rb
+++ b/challenge-113/abigail/ruby/ch-1.rb
@@ -8,20 +8,30 @@
# Run as: ruby ch-1.rb < input-file
#
-tens = [0, 0, 1, 2, 1, 0, 2, 6, 3, 8]
+#
+# For a description of the algorithm, and the proof why this is correct:
+# https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-113-1.html
+#
+
+gcds = [0, 1, 2, 1, 2, 5, 2, 1, 2, 1]
ARGF . each_line do
| line |
n, d = line . split
n = n . to_i
d = d . to_i
- d10 = d == 0 ? 100 : d * 10
- if n >= d10 || n % (d == 0 ? 10 : d) == 0
+ if d == 0
+ puts (n >= 100 || n % 10 == 0 ? 1 : 0)
+ next
+ end
+
+ if n >= d * 10
then puts (1)
next
end
+
done = false
- for i in 1 .. tens [d] do
+ for i in 0 .. d / gcds [d] - 1 do
t = n - 10 * i - d
if t >= 0 && t % d == 0
then puts (1)
generated by cgit (git 2.34.1) at 2026-07-13 19:47:04 +0000