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Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com> Co-authored-by: Christian Clauss <cclauss@me.com>
58 lines
1.5 KiB
Python
58 lines
1.5 KiB
Python
"""
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Problem 125: https://projecteuler.net/problem=125
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The palindromic number 595 is interesting because it can be written as the sum
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of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.
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There are exactly eleven palindromes below one-thousand that can be written as
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consecutive square sums, and the sum of these palindromes is 4164. Note that
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1 = 0^2 + 1^2 has not been included as this problem is concerned with the
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squares of positive integers.
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Find the sum of all the numbers less than 10^8 that are both palindromic and can
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be written as the sum of consecutive squares.
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"""
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LIMIT = 10**8
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def is_palindrome(n: int) -> bool:
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"""
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Check if an integer is palindromic.
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>>> is_palindrome(12521)
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True
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>>> is_palindrome(12522)
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False
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>>> is_palindrome(12210)
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False
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"""
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if n % 10 == 0:
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return False
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s = str(n)
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return s == s[::-1]
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def solution() -> int:
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"""
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Returns the sum of all numbers less than 1e8 that are both palindromic and
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can be written as the sum of consecutive squares.
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"""
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answer = set()
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first_square = 1
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sum_squares = 5
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while sum_squares < LIMIT:
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last_square = first_square + 1
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while sum_squares < LIMIT:
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if is_palindrome(sum_squares):
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answer.add(sum_squares)
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last_square += 1
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sum_squares += last_square**2
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first_square += 1
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sum_squares = first_square**2 + (first_square + 1) ** 2
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return sum(answer)
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if __name__ == "__main__":
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print(solution())
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