示例#1
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def compute():
    # Consider an arbitrary fraction n/d:
    #   Let n = 10 * n1 + n0 be the numerator.
    #   Let d = 10 * d1 + d0 be the denominator.
    # As stated in the problem, we need 10 <= n < d < 100.
    # We must disregard trivial simplifications where n0 = d0 = 0.
    #
    # Now, a simplification with n0 = d0 is impossible because:
    #   n1 / d1 = n / d = (10*n1 + n0) / (10*d1 + n0).
    #   n1 * (10*d1 + n0) = d1 * (10*n1 + n0).
    #   10*n1*d1 + n1*n0 = 10*d1*n1 + d1*n0.
    #   n1*n0 = d1*n0.
    #   n1 = d1.
    #   This implies n = d, which contradicts the fact that n < d.
    # Similarly, we cannot have a simplification with n1 = d1 for the same reason.
    #
    # Therefore we only need to consider the cases where n0 = d1 or n1 = d0.
    # In the first case, check that n1/d0 = n/d;
    # in the second case, check that n0/d1 = n/d.
    numer = 1
    denom = 1
    for d in range(10, 100):
        for n in range(10, d):
            n0 = n % 10
            n1 = n // 10
            d0 = d % 10
            d1 = d // 10
            if (n1 == d0 and n0 * d == n * d1) or (n0 == d1 and n1 * d == n * d0):
                numer *= n
                denom *= d
    return str(denom // eulerlib.gcd(numer, denom))
def compute():
    # Consider an arbitrary fraction n/d:
    #   Let n = 10 * n1 + n0 be the numerator.
    #   Let d = 10 * d1 + d0 be the denominator.
    # As stated in the problem, we need 10 <= n < d < 100.
    # We must disregard trivial simplifications where n0 = d0 = 0.
    #
    # Now, a simplification with n0 = d0 is impossible because:
    #   n1 / d1 = n / d = (10*n1 + n0) / (10*d1 + n0).
    #   n1 * (10*d1 + n0) = d1 * (10*n1 + n0).
    #   10*n1*d1 + n1*n0 = 10*d1*n1 + d1*n0.
    #   n1*n0 = d1*n0.
    #   n1 = d1.
    #   This implies n = d, which contradicts the fact that n < d.
    # Similarly, we cannot have a simplification with n1 = d1 for the same reason.
    #
    # Therefore we only need to consider the cases where n0 = d1 or n1 = d0.
    # In the first case, check that n1/d0 = n/d;
    # in the second case, check that n0/d1 = n/d.
    numer = 1
    denom = 1
    for d in range(10, 100):
        for n in range(10, d):
            n0 = n % 10
            n1 = n // 10
            d0 = d % 10
            d1 = d // 10
            if (n1 == d0 and n0 * d == n * d1) or (n0 == d1
                                                   and n1 * d == n * d0):
                numer *= n
                denom *= d
    return str(denom // eulerlib.gcd(numer, denom))
def is_prime(n, k):
    """
    Test if a number n is prime k-times.
    :param n: The prime number to be tested.
    :param k: The number of tests.
    :return:
    """
    if n <= 1 or n == 4:
        return False
    if n <= 3:
        return True
    if eulerlib.is_even(n):
        return False
    while k > 0:

        # Take random int in [2, n-2]
        a = random.randint(2, n-1)

        # Check if a and n are co-prime.
        if eulerlib.gcd(n, a) != 1:
            return False

        # Fermat's little theorem
        if modpow(a, n-1, n) != 1:
            return False

        k -= 1

    return True
def digit_cancelling_fractions():
    lowest_nums = []
    lowest_denoms = []
    for den in range(10, 100):
        if den % 10 == 0:
            continue
        for num in range(10, den):
            num_digits = eulerlib.digits(num)
            den_digits = eulerlib.digits(den)
            if num % 10 == 0:
                continue
            digits_intersection = list(set(num_digits) & set(den_digits))
            if len(digits_intersection) > 0:
                num_digits.remove(digits_intersection[0])
                den_digits.remove(digits_intersection[0])
                num_lct = eulerlib.digits_to_int(num_digits)
                den_lct = eulerlib.digits_to_int(den_digits)
                cancelled_fraction = num_lct / den_lct
                fraction = num / den
                if fraction == cancelled_fraction:
                    lowest_nums.append(num_lct)
                    lowest_denoms.append(den_lct)
                    print('{} / {} = {} / {}'.format(num, den, num_lct, den_lct))

    num_prod = eulerlib.product(lowest_nums)
    den_prod = eulerlib.product(lowest_denoms)
    gcd = eulerlib.gcd(num_prod, den_prod)
    return den_prod / gcd
def count_all_unconcealed(prime):
	result = []
	for e in range(prime - 1):
		if eulerlib.gcd(e, prime - 1) == 1:
			result.append(count_unconcealed(prime, e))
		else:
			result.append(10**20)  # Sentinel
	return result
示例#6
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def solve():
    qtys = [0] * (bound+1)
    for m in xrange(2, int(ceil((sqrt(1+2*bound) - 1)/2))+1):
        for n in xrange(2 - (m+1)%2, m, 2):
            if gcd(m,n) != 1: continue
            perim = 2*m*(m+n)
            if perim > bound: break
            for x in xrange(perim, bound+1, perim):
                qtys[x] += 1
    return sum(1 for q in qtys if q == 1)
示例#7
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def solve():
    for a in range(1, 10):
        for b in range(1, 10):
            ab = a*10 + b
            for c in range(1, 10):
                checkFrac(ab,b,c, a,c)  # ab/bc = a/c
                checkFrac(ab,c,b, a,c)  # ab/cb = a/c
                checkFrac(ab,a,c, b,c)  # ab/ac = b/c
                checkFrac(ab,c,a, b,c)  # ab/ca = b/c
    (nums, denoms) = zip(*fractions)
    num = product(nums)
    denom = product(denoms)
    return denom // gcd(num, denom)
示例#8
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def find(n: int):
    p = 1
    q = 1
    for a in range(1, n):
        for b in range(1, n):
            for c in range(1, n):
                if f(a, b, c):
                    p *= (10 * a + b)
                    q *= (10 * b + c)
                elif g(a, b, c):
                    p *= (10 * a + b)
                    q *= (10 * c + b)
    r = lib.gcd(p, q)
    p //= r
    q //= r
    return (p, q)
示例#9
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def getPnum(edgeMAX):
    res = {}
    m = 2
    flag = True
    while flag:
        for n in range(m - 1, 0, -2):
            if m * m + n * n <= edgeMAX:
                if gcd(m, n) == 1:
                    if m * m + n * n in res:
                        res[m * m + n * n].append((2 * m * n, m * m - n * n))
                    else:
                        res[m * m + n * n] = [(2 * m * n, m * m - n * n)]
            else:
                flag = False
                break
        m += 1
    return res
示例#10
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def getPnumAll(edgeMAX):
    res = {}
    m = 2
    flag = True
    while flag:
        for n in range(m - 1, 0, -2):
            if m * m + n * n <= edgeMAX:
                if gcd(m, n) == 1:
                    t = edgeMAX // (m * m + n * n)
                    for i in range(1, t + 1):
                        if (m * m + n * n) * i in res:
                            res[(m * m + n * n) * i].append(
                                (2 * m * n * i, (m * m - n * n) * i))
                        else:
                            res[(m * m + n * n) * i] = [(2 * m * n * i,
                                                         (m * m - n * n) * i)]
            else:
                flag = False
                break
        m += 1
    return res
示例#11
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 def simplify(self):
     c = eulerlib.gcd(self.num, self.den)
     return Fraction(self.num // c, self.den // c)
示例#12
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def compute():
    ans = 1
    for i in range(1, 21):
        ans *= i // eulerlib.gcd(i, ans)
    return str(ans)
示例#13
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from eulerlib import cmpf, gcd

N = 12000 

f1 = (1,3)
f2 = (1,2)

c = 0
for d in xrange(1,N+1):
	for n in xrange(d/3,d/2+1):
		if gcd(n,d) == 1:
			f = (n,d)
			if cmpf(f,f1) > 0 and cmpf(f,f2) < 0:
				c += 1

print c
示例#14
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def totient(n):
    result = 0
    for k in xrange(1, n + 1):
        if gcd(k, n) == 1:
            result += 1
    return result
示例#15
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def compute():
    ans = 1
    for i in range(1, 21):
        ans *= i // eulerlib.gcd(i, ans)
    return str(ans)