41 lines
879 B
Python
41 lines
879 B
Python
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from lib_prime import prime_nth
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def varray(i):
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return i * 2 + 1
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def rem(i):
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n = varray(i)
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p = prime_nth(n)
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return (pow(p - 1, n) + pow(p + 1, n)) % (p * p)
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def bin_search(lo, hi, target):
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if hi - lo == 1:
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return varray(hi)
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i = lo + ((hi - lo) // 2)
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r = rem(i)
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# print(f"{i=:<6} {r=}")
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if r < target:
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return bin_search(i, hi, target)
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elif r > target:
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return bin_search(lo, i, target)
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assert False
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def euler_123():
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"""
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I found out that when n is even, the remainder is always 2. When the
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remainder is odd, the series grows monotonically. That means we can use a
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binary search on odd values to find the solution.
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"""
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return bin_search(1000, 30000, 10**10)
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if __name__ == "__main__":
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solution = euler_123()
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print("e123.py: " + str(solution))
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assert(solution == 21035)
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