From e64734f9b42c5838a7d3ade2d6c2e8ba14625cf0 Mon Sep 17 00:00:00 2001
From: Paulo Custodio <pauloscustodio@gmail.com>
Date: Tue, 9 Mar 2021 19:39:28 +0000
Subject: Add Lua and Python solutions to challenge 102

---
 challenge-102/paulo-custodio/python/ch-2.py | 44 +++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 challenge-102/paulo-custodio/python/ch-2.py

(limited to 'challenge-102/paulo-custodio/python/ch-2.py')

diff --git a/challenge-102/paulo-custodio/python/ch-2.py b/challenge-102/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..b996cf5522
--- /dev/null
+++ b/challenge-102/paulo-custodio/python/ch-2.py
@@ -0,0 +1,44 @@
+#!/usr/bin/env python
+
+# Challenge 102
+# 
+# TASK #2 > Hash-counting String
+# Submitted by: Stuart Little
+# 
+# You are given a positive integer $N.
+# 
+# Write a script to produce Hash-counting string of that length.
+# 
+# The definition of a hash-counting string is as follows:
+# - the string consists only of digits 0-9 and hashes, '#'
+# - there are no two consecutive hashes: '##' does not appear in your string
+# - the last character is a hash
+# - the number immediately preceding each hash (if it exists) is the position 
+# of that hash in the string, with the position being counted up from 1
+# 
+# It can be shown that for every positive integer N there is exactly one such 
+# length-N string.
+# Examples:
+# 
+# (a) "#" is the counting string of length 1
+# (b) "2#" is the counting string of length 2
+# (c) "#3#" is the string of length 3
+# (d) "#3#5#7#10#" is the string of length 10
+# (e) "2#4#6#8#11#14#" is the string of length 14
+
+import sys
+
+def hash_counting(n):
+	out = ''
+	i = n
+	while i>0:
+		p = i
+		out = '#'+out
+		i -= 1
+		while i>0 and p!=0:
+			out = str(p % 10)+out
+			i -= 1
+			p = int(p/10)
+	return out
+
+print(hash_counting(int(sys.argv[1])))
-- 
cgit 


From c51f9bd979dbd79f20cb0abd543dc9b4df9f4e84 Mon Sep 17 00:00:00 2001
From: Paulo Custodio <pauloscustodio@gmail.com>
Date: Tue, 9 Mar 2021 20:01:18 +0000
Subject: Remove tabs

---
 challenge-102/paulo-custodio/python/ch-2.py | 38 ++++++++++++++---------------
 1 file changed, 19 insertions(+), 19 deletions(-)

(limited to 'challenge-102/paulo-custodio/python/ch-2.py')

diff --git a/challenge-102/paulo-custodio/python/ch-2.py b/challenge-102/paulo-custodio/python/ch-2.py
index b996cf5522..558367a422 100644
--- a/challenge-102/paulo-custodio/python/ch-2.py
+++ b/challenge-102/paulo-custodio/python/ch-2.py
@@ -1,25 +1,25 @@
 #!/usr/bin/env python
 
 # Challenge 102
-# 
+#
 # TASK #2 > Hash-counting String
 # Submitted by: Stuart Little
-# 
+#
 # You are given a positive integer $N.
-# 
+#
 # Write a script to produce Hash-counting string of that length.
-# 
+#
 # The definition of a hash-counting string is as follows:
 # - the string consists only of digits 0-9 and hashes, '#'
 # - there are no two consecutive hashes: '##' does not appear in your string
 # - the last character is a hash
-# - the number immediately preceding each hash (if it exists) is the position 
+# - the number immediately preceding each hash (if it exists) is the position
 # of that hash in the string, with the position being counted up from 1
-# 
-# It can be shown that for every positive integer N there is exactly one such 
+#
+# It can be shown that for every positive integer N there is exactly one such
 # length-N string.
 # Examples:
-# 
+#
 # (a) "#" is the counting string of length 1
 # (b) "2#" is the counting string of length 2
 # (c) "#3#" is the string of length 3
@@ -29,16 +29,16 @@
 import sys
 
 def hash_counting(n):
-	out = ''
-	i = n
-	while i>0:
-		p = i
-		out = '#'+out
-		i -= 1
-		while i>0 and p!=0:
-			out = str(p % 10)+out
-			i -= 1
-			p = int(p/10)
-	return out
+    out = ''
+    i = n
+    while i>0:
+        p = i
+        out = '#'+out
+        i -= 1
+        while i>0 and p!=0:
+            out = str(p % 10)+out
+            i -= 1
+            p = int(p/10)
+    return out
 
 print(hash_counting(int(sys.argv[1])))
-- 
cgit