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| author | Dave Jacoby <jacoby.david@gmail.com> | 2024-09-18 19:56:42 -0400 |
|---|---|---|
| committer | Dave Jacoby <jacoby.david@gmail.com> | 2024-09-18 19:56:42 -0400 |
| commit | f86f5e2fec16020c1d86f9028fb0f61cfeac106e (patch) | |
| tree | 0fd388a696b51ffde5a7bfe8519a74e1caf42461 /challenge-059/paulo-custodio/python/ch-2.py | |
| parent | ff8719c86653d5ad3121955e9494a0010527c2b9 (diff) | |
| parent | 0052ec63ca70eaa6d9ffb1926c294dbfd85f8c05 (diff) | |
| download | perlweeklychallenge-club-f86f5e2fec16020c1d86f9028fb0f61cfeac106e.tar.gz perlweeklychallenge-club-f86f5e2fec16020c1d86f9028fb0f61cfeac106e.tar.bz2 perlweeklychallenge-club-f86f5e2fec16020c1d86f9028fb0f61cfeac106e.zip | |
Merge branch 'master' of https://github.com/manwar/perlweeklychallenge-club
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_) |
