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44 lines
1.1 KiB
Python
44 lines
1.1 KiB
Python
"""
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This program calculates the nth Fibonacci number in O(log(n)).
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It's possible to calculate F(1000000) in less than a second.
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"""
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from __future__ import print_function
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import sys
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# returns F(n)
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def fibonacci(n: int): # noqa: F821 This syntax is Python 3 only
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if n < 0:
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raise ValueError("Negative arguments are not supported")
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return _fib(n)[0]
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# returns (F(n), F(n-1))
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def _fib(n: int): # noqa: F821 This syntax is Python 3 only
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if n == 0:
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# (F(0), F(1))
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return (0, 1)
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else:
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# F(2n) = F(n)[2F(n+1) − F(n)]
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# F(2n+1) = F(n+1)^2+F(n)^2
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a, b = _fib(n // 2)
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c = a * (b * 2 - a)
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d = a * a + b * b
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if n % 2 == 0:
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return (c, d)
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else:
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return (d, c + d)
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if __name__ == "__main__":
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args = sys.argv[1:]
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if len(args) != 1:
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print("Too few or too much parameters given.")
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exit(1)
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try:
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n = int(args[0])
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except ValueError:
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print("Could not convert data to an integer.")
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exit(1)
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print("F(%d) = %d" % (n, fibonacci(n)))
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