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* ci(pre-commit): Add pep8-naming to `pre-commit` hooks (#7038) * refactor: Fix naming conventions (#7038) * Update arithmetic_analysis/lu_decomposition.py Co-authored-by: Christian Clauss <cclauss@me.com> * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * refactor(lu_decomposition): Replace `NDArray` with `ArrayLike` (#7038) * chore: Fix naming conventions in doctests (#7038) * fix: Temporarily disable project euler problem 104 (#7069) * chore: Fix naming conventions in doctests (#7038) Co-authored-by: Christian Clauss <cclauss@me.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
51 lines
1.2 KiB
Python
51 lines
1.2 KiB
Python
"""
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Consider all integer combinations of ab for 2 <= a <= 5 and 2 <= b <= 5:
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2^2=4, 2^3=8, 2^4=16, 2^5=32
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3^2=9, 3^3=27, 3^4=81, 3^5=243
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4^2=16, 4^3=64, 4^4=256, 4^5=1024
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5^2=25, 5^3=125, 5^4=625, 5^5=3125
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If they are then placed in numerical order, with any repeats removed, we get
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the following sequence of 15 distinct terms:
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4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
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How many distinct terms are in the sequence generated by ab
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for 2 <= a <= 100 and 2 <= b <= 100?
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"""
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def solution(n: int = 100) -> int:
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"""Returns the number of distinct terms in the sequence generated by a^b
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for 2 <= a <= 100 and 2 <= b <= 100.
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>>> solution(100)
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9183
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>>> solution(50)
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2184
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>>> solution(20)
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324
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>>> solution(5)
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15
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>>> solution(2)
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1
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>>> solution(1)
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0
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"""
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collect_powers = set()
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current_pow = 0
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n = n + 1 # maximum limit
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for a in range(2, n):
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for b in range(2, n):
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current_pow = a**b # calculates the current power
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collect_powers.add(current_pow) # adds the result to the set
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return len(collect_powers)
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if __name__ == "__main__":
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print("Number of terms ", solution(int(str(input()).strip())))
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