dlog.py 文件源码

def chinese_remainder(n, a):
    ''' Compute a number from its moduluses.

    From Rosetta Code. '''

    sum = 0
    prod = reduce(lambda a, b: a*b, n)

    for n_i, a_i in zip(n, a):
        p = prod / n_i
        sum += a_i * number.inverse(p, n_i) * p
    return sum % prod