"""
Problem 33: https://projecteuler.net/problem=33

The fraction 49/98 is a curious fraction, as an inexperienced
mathematician in attempting to simplify it may incorrectly believe
that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction,
less than one in value, and containing two digits in the numerator
and denominator.

If the product of these four fractions is given in its lowest common
terms, find the value of the denominator.
"""

# _*_ conding:UTF-8 _*_
'''
@author = Kuperain
@email = kuperain@aliyun.com
@IDE = VSCODE Python3.8.3
@creat_time = 2022/5/12
'''


def solution1() -> int:
    '''
    in the case d=2(two digits in the numerator and denominator)
        (ab)/(cd), where 
            (ab) < (cd), 1 <= a <= c
            if a==c, (ab)/(ad) == b/d, => b=d  X
            if b==d, (ab)/(cb) == a/c, => a=c  X
            if a==d, (ab)/(ca) == b/c, => a(10c-b) = 9bc
            if b==c, (ab)/(bd) == a/d, => b(10a-d) = 9ad

    more generally, (a1,a2,a2,...ad)/(b1,b2,b2,...bd),
    we can rule out two cases:
        a1 = b1  X
        ad = bd  X

    '''

    finds = list()

    for m in range(10, 99):
        if m % 10 == 0:
            continue
        for n in range(m+1, 99):
            ms = str(m)
            ns = str(n)
            if len(set(ms + ns)) == 4:
                # print(f'{m},{n} do not have same digit.')
                continue

            for i in range(2):
                for j in range(2):
                    if ms[i] == ns[j]:
                        # print(f'{m},{n},both have {ms[i]}.')

                        ms1 = list(ms)
                        ms1.remove(ms[i])
                        m1 = int(''.join(ms1))

                        ns1 = list(ns)
                        ns1.remove(ns[j])
                        n1 = int(''.join(ns1))

                        if n1 != 0 and m*n1 == n*m1:
                            finds.append((m, n))

    print(finds)
    # [(16, 64), (19, 95), (26, 65), (49, 98)]

    p, q = 1, 1
    for fract in finds:
        p *= fract[0]
        q *= fract[1]

    # print(p, q)
    # 387296 38729600

    import math
    return q//math.gcd(p, q)


def solution2(dm: int = 2, dn: int = 2) -> int:
    '''
    (a1,a2,a2,...a_dm)/(b1,b2,b2,...b_dn)
    '''

    if dm > dn:
        raise ValueError('m < n')

    finds = set()

    for m in range(10**(dm-1), 10**dm):
        for n in range(max(m+1, 10**(dn-1)), 10**dn):
            # dm=dm-1, dn=dn-1, not belong  (dm,dn) case
            if m % 10 == 0 or n % 10 == 0:
                continue
            ms = str(m)
            ns = str(n)
            for i in range(dm):
                if i < dm-1 and ms[i+1] == ms[i]:
                    continue
                for j in range(dn):
                    if j < dn-1 and ns[j+1] == ns[j]:
                        continue
                    if ms[i] == ns[j]:
                        msl = list(ms)
                        msl.pop(i)
                        msl = int(''.join(msl))

                        nsl = list(ns)
                        nsl.pop(j)
                        nsl = int(''.join(nsl))

                        if m*nsl == n*msl:
                            if (m, n) in finds:
                                print(m, n)
                            finds.add((m, n, int(ms[i])))
    print(len(finds))
    return finds


if __name__ == "__main__":
    import doctest
    doctest.testmod(verbose=False)

    print(solution1())
    # 100
    solution2(3, 3)
    # 245