Created
December 14, 2013 12:35
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check integer divisible by 3 using alternative sum
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#! /usr/bin/env python | |
# -*- coding: utf-8 -*- | |
# vim:fenc=utf-8 | |
# | |
# Copyright © 2013 KuoE0 <kuoe0.tw@gmail.com> | |
# | |
# Distributed under terms of the MIT license. | |
""" | |
""" | |
def alt_add_digits_recursion(x, sign): | |
if x > 1: | |
return alt_add_digits(x >> 1, sign * -1) + (x & 1) * sign | |
return (x & 1) * sign | |
def alt_add_digits_iteration(x): | |
ret = 0 | |
sign = 1 | |
while x > 1: | |
ret = ret + (x & 1) * sign | |
x = x >> 1 | |
sign = sign * -1 | |
return ret | |
def alt_add_digits_iteration_no_mul(x): | |
ret = 0 | |
sign = True | |
while x > 1: | |
ret = ret + (x & 1) if sign else ret - (x & 1) | |
x = x >> 1 | |
sign = not sign | |
return ret | |
def divisible_3(x): | |
while x > 1: | |
x = alt_add_digits_iteration_no_mul(x) | |
x = x if x >= 0 else -x | |
return x == 0 | |
if __name__ == "__main__": | |
import sys | |
for i in range(int(sys.argv[1])): | |
print i, divisible_3(i) | |
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