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| author | Paulo Custodio <pauloscustodio@gmail.com> | 2024-09-15 13:03:59 +0100 |
|---|---|---|
| committer | Paulo Custodio <pauloscustodio@gmail.com> | 2024-09-15 13:03:59 +0100 |
| commit | 1e1739da53b6ecbe5f17fa6cf3b922ad6ff5d347 (patch) | |
| tree | b6c71eae46bba9915a707687817c4acbf9e3ce24 /challenge-059/paulo-custodio/python/ch-2.py | |
| parent | e806566dab3ce64c5f5498b25bfd0dc52de51709 (diff) | |
| download | perlweeklychallenge-club-1e1739da53b6ecbe5f17fa6cf3b922ad6ff5d347.tar.gz perlweeklychallenge-club-1e1739da53b6ecbe5f17fa6cf3b922ad6ff5d347.tar.bz2 perlweeklychallenge-club-1e1739da53b6ecbe5f17fa6cf3b922ad6ff5d347.zip | |
Add Python solution to challenge 059
Diffstat (limited to 'challenge-059/paulo-custodio/python/ch-2.py')
| -rw-r--r-- | challenge-059/paulo-custodio/python/ch-2.py | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/challenge-059/paulo-custodio/python/ch-2.py b/challenge-059/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..89d4f6741d --- /dev/null +++ b/challenge-059/paulo-custodio/python/ch-2.py @@ -0,0 +1,42 @@ +#!/usr/bin/env python3 + +# Challenge 059 +# +# TASK #2 > Bit Sum +# Reviewed by Ryan Thompson +# Helper Function +# For this task, you will most likely need a function f(a,b) which returns the +# count of different bits of binary representation of a and b. +# +# For example, f(1,3) = 1, since: +# +# Binary representation of 1 = 01 +# +# Binary representation of 3 = 11 +# +# There is only 1 different bit. Therefore the subroutine should return 1. Note +# that if one number is longer than the other in binary, the most significant +# bits of the smaller number are padded (i.e., they are assumed to be zeroes). +# +# Script Output +# You script should accept n positive numbers. Your script should sum the result +# of f(a,b) for every pair of numbers given: +# +# For example, given 2, 3, 4, the output would be 6, +# since f(2,3) + f(2,4) + f(3,4) = 1 + 2 + 3 = 6 + +import sys +from itertools import combinations + +def f(a, b): + r = int(a) ^ int(b) + rt = bin(r) + return rt.count('1') + +n = list(map(int, sys.argv[1:])) +sum_ = 0 + +for combin in combinations(n, 2): + sum_ += f(combin[0], combin[1]) + +print(sum_) |
