예제 #1
0
def euler026(maxd):
    ''' A unit fraction contains 1 in the numerator. 
    
    The decimal representation of the unit fractions with denominators 2 to 10 are given
    1/2 = 0.5
    1/3 = 0.(3)
    1/4 = 0.25
    1/5 = 0.2
    1/6 = 0.1(6)
    1/7 = 0.(142857)
    1/8 = 0.125
    1/9 = 0.(1)
    1/10 = 0.1

    Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

    Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
    
    >>> euler026(10)
    7
    >>> euler026(1000)
    983
    '''
    b = Biggest()
    for p in primes(maxd):
        b.set(find_repeat(p), p)
    return b.data
예제 #2
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def euler035(limit):
    """ Circular primes
    
    The number, 197, is called a circular prime because all rotations of 
    the digits: 197, 971, and 719, are themselves prime.

    There are thirteen such primes below 100: 
    
    2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

    How many circular primes are there below one million?

    >>> euler035(100)
    13
    >>> euler035(1000000)
    55
    """
    c = count()
    P = set(primes(limit))

    for i in P:
        if all(r in P for r in rotate(i)):
            c.next()

    return c.next()
예제 #3
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def euler027(limit):
    '''Quadratic primes
    
    Euler discovered the remarkable quadratic formula:

    n**2 + n + 41
    
    It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39.
    However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when
    n = 41, 41n**2 + 41 + 41 is clearly divisible by 41.

    The incredible formula  n**2 - 79n + 1601 was discovered, which produces 80 primes for the
    consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.

    Considering quadratics of the form:

    n**2 + an + b, where |a| < 1000 and |b| < 1000

    where |n| is the modulus/absolute value of n
    e.g. |11| = 11 and |-4| = 4
    Find the product of the coefficients, a and b, for the quadratic expression that produces
    the maximum number of primes for consecutive values of n, starting with n = 0.
    '''
    P = Primes()
    B = Biggest()
    for a, b in product(range(-limit+1, limit), primes(limit)):
        B.set(len(list(takewhile(lambda x: quadratic(x, a, b) in P, count()))), a * b)
    return B.data