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author冯昶 <fengchang@novel-supertv.com>2024-09-09 15:10:21 +0800
committer冯昶 <fengchang@novel-supertv.com>2024-09-09 15:10:21 +0800
commit90e81fa7a4d4ba2eb482542cde3401f8c166adc7 (patch)
tree32574bfed25f9181a5281395f5dc3dd0321e0de4 /challenge-285/pokgopun/python
parentd7a3db86265e08657df10663e50d63d87d6695d1 (diff)
parent3c67a5382758155040d6598a2fa01ca5fd6d25d9 (diff)
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Merge remote-tracking branch 'upstream/master'
Diffstat (limited to 'challenge-285/pokgopun/python')
-rw-r--r--challenge-285/pokgopun/python/ch-1.py54
-rw-r--r--challenge-285/pokgopun/python/ch-2.py72
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diff --git a/challenge-285/pokgopun/python/ch-1.py b/challenge-285/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-285/
+"""
+
+Task 1: No Connection
+
+Submitted by: [49]Mohammad Sajid Anwar
+ __________________________________________________________________
+
+ You are given a list of routes, @routes.
+
+ Write a script to find the destination with no further outgoing
+ connection.
+
+Example 1
+
+Input: @routes = (["B","C"], ["D","B"], ["C","A"])
+Output: "A"
+
+"D" -> "B" -> "C" -> "A".
+"B" -> "C" -> "A".
+"C" -> "A".
+"A".
+
+Example 2
+
+Input: @routes = (["A","Z"])
+Output: "Z"
+
+Task 2: Making Change
+"""
+### solution by pokgopun@gmail.com
+
+def nc(routes: tuple):
+ dct = dict()
+ for src,dst in routes:
+ if dct.get(src,True):
+ dct[src] = False
+ dct.setdefault(dst,True)
+ for k,v in dct.items():
+ if v:
+ return k
+ return None
+
+import unittest
+
+class TestNc(unittest.TestCase):
+ def test(self):
+ for inpt, otpt in {
+ (("B","C"), ("D","B"), ("C","A")): "A",
+ (("A","Z"),): "Z",
+ }.items():
+ self.assertEqual(nc(inpt),otpt)
+
+unittest.main()
diff --git a/challenge-285/pokgopun/python/ch-2.py b/challenge-285/pokgopun/python/ch-2.py
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+++ b/challenge-285/pokgopun/python/ch-2.py
@@ -0,0 +1,72 @@
+### https://theweeklychallenge.org/blog/perl-weekly-challenge-285/
+"""
+
+Task 2: Making Change
+
+Submitted by: [50]David Ferrone
+ __________________________________________________________________
+
+ Compute the number of ways to make change for given amount in cents. By
+ using the coins e.g. Penny, Nickel, Dime, Quarter and Half-dollar, in
+ how many distinct ways can the total value equal to the given amount?
+ Order of coin selection does not matter.
+A penny (P) is equal to 1 cent.
+A nickel (N) is equal to 5 cents.
+A dime (D) is equal to 10 cents.
+A quarter (Q) is equal to 25 cents.
+A half-dollar (HD) is equal to 50 cents.
+
+Example 1
+
+Input: $amount = 9
+Ouput: 2
+
+1: 9P
+2: N + 4P
+
+Example 2
+
+Input: $amount = 15
+Ouput: 6
+
+1: D + 5P
+2: D + N
+3: 3N
+4: 2N + 5P
+5: N + 10P
+6: 15P
+
+Example 3
+
+Input: $amount = 100
+Ouput: 292
+ __________________________________________________________________
+
+ Last date to submit the solution 23:59 (UK Time) Sunday 8th September
+ 2024.
+ __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def mc(amount: int):
+ c = 0
+ for a50 in range(0,amount+1,50):
+ for a25 in range(0,amount-a50+1,25):
+ for a10 in range(0,amount-a50-a25+1,10):
+ c += (amount-a50-a25-a10)//5 + 1
+ return c
+
+import unittest
+
+class TestMc(unittest.TestCase):
+ def test(self):
+ for inpt, otpt in {
+ 9: 2,
+ 15: 6,
+ 100: 292,
+ }.items():
+ self.assertEqual(mc(inpt),otpt)
+
+unittest.main()