Daniele Fucini
a0c155d58c
Added solution for problems 102 and 124 in C, for problem 104 in python and for problem 145 in both C and python.
39 lines
1.1 KiB
Python
39 lines
1.1 KiB
Python
#!/usr/bin/python
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# Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits.
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# For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
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# Leading zeroes are not allowed in either n or reverse(n).
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#
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# There are 120 reversible numbers below one-thousand.
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#
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# How many reversible numbers are there below one-billion (109)?
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from timeit import default_timer
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def main():
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start = default_timer()
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N = 10000000
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count = 0
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# Brute force approach, sum each number and their reverse and
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# check if there are only odd digits.
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for i in range(11, N):
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if i % 10 != 0:
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s = str(i + int(''.join(reversed(str(i)))))
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if not '0' in s and not '2' in s and not '4' in s and\
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not '6' in s and not '8' in s:
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count = count + 1
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end = default_timer()
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print('Project Euler, Problem 145')
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print('Answer: {}'.format(count))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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