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142 lines
2.9 KiB
Python
142 lines
2.9 KiB
Python
"""
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Project Euler Problem 104 : https://projecteuler.net/problem=104
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The Fibonacci sequence is defined by the recurrence relation:
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Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
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It turns out that F541, which contains 113 digits, is the first Fibonacci number
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for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9,
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but not necessarily in order). And F2749, which contains 575 digits, is the first
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Fibonacci number for which the first nine digits are 1-9 pandigital.
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Given that Fk is the first Fibonacci number for which the first nine digits AND
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the last nine digits are 1-9 pandigital, find k.
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"""
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import sys
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sys.set_int_max_str_digits(0) # type: ignore
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def check(number: int) -> bool:
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"""
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Takes a number and checks if it is pandigital both from start and end
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>>> check(123456789987654321)
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True
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>>> check(120000987654321)
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False
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>>> check(1234567895765677987654321)
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True
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"""
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check_last = [0] * 11
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check_front = [0] * 11
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# mark last 9 numbers
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for _ in range(9):
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check_last[int(number % 10)] = 1
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number = number // 10
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# flag
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f = True
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# check last 9 numbers for pandigitality
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for x in range(9):
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if not check_last[x + 1]:
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f = False
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if not f:
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return f
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# mark first 9 numbers
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number = int(str(number)[:9])
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for _ in range(9):
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check_front[int(number % 10)] = 1
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number = number // 10
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# check first 9 numbers for pandigitality
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for x in range(9):
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if not check_front[x + 1]:
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f = False
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return f
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def check1(number: int) -> bool:
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"""
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Takes a number and checks if it is pandigital from END
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>>> check1(123456789987654321)
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True
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>>> check1(120000987654321)
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True
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>>> check1(12345678957656779870004321)
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False
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"""
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check_last = [0] * 11
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# mark last 9 numbers
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for _ in range(9):
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check_last[int(number % 10)] = 1
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number = number // 10
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# flag
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f = True
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# check last 9 numbers for pandigitality
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for x in range(9):
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if not check_last[x + 1]:
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f = False
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return f
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def solution() -> int:
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"""
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Outputs the answer is the least Fibonacci number pandigital from both sides.
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>>> solution()
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329468
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"""
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a = 1
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b = 1
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c = 2
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# temporary Fibonacci numbers
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a1 = 1
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b1 = 1
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c1 = 2
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# temporary Fibonacci numbers mod 1e9
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# mod m=1e9, done for fast optimisation
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tocheck = [0] * 1000000
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m = 1000000000
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for x in range(1000000):
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c1 = (a1 + b1) % m
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a1 = b1 % m
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b1 = c1 % m
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if check1(b1):
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tocheck[x + 3] = 1
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for x in range(1000000):
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c = a + b
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a = b
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b = c
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# perform check only if in tocheck
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if tocheck[x + 3] and check(b):
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return x + 3 # first 2 already done
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return -1
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if __name__ == "__main__":
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print(f"{solution() = }")
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