Daniele Fucini
a0c155d58c
Added solution for problems 102 and 124 in C, for problem 104 in python and for problem 145 in both C and python.
58 lines
1.8 KiB
Python
58 lines
1.8 KiB
Python
# The Fibonacci sequence is defined by the recurrence relation:
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#
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# F_n = F_n−1 + F_n−2, where F_1 = 1 and F_2 = 1.
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# It turns out that F_541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital
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# (contain all the digits 1 to 9, but not necessarily in order). And F_2749, which contains 575 digits, is the first Fibonacci number
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# for which the first nine digits are 1-9 pandigital.
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#
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# Given that F_k is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
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#!/usr/bin/python
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from mpmath import matrix, mp
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from timeit import default_timer
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from projecteuler import is_pandigital
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def main():
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start = default_timer()
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found = 0
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fib1 = 1
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fib2 = 1
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i = 2
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while not found:
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# Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.
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fibn = (fib1 + fib2) % 1000000000
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fib1 = fib2
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fib2 = fibn
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i = i + 1
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# If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
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# the matrix representation, with sufficient precision so that the at least the first
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# 9 digits are correct (we don't need the whole number. If the first 9 digits are also
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# pandigital, we found the solution.
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if is_pandigital(fibn, 9):
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fib_matrix = matrix(2)
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fib_matrix[0, 0] = 1
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fib_matrix[0, 1] = 1
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fib_matrix[1, 0] = 1
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fib_matrix[1, 1] = 0
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fib = fib_matrix ** i
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fib = int(fib[0, 1])
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if is_pandigital(int(str(fib)[:9]), 9):
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found = 1
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end = default_timer()
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print('Project Euler, Problem 104')
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print('Answer: {}'.format(i))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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