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Home/ Questions/Q 6034001
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T05:33:59+00:00 2026-05-23T05:33:59+00:00

Is there a more efficient way to write this so it’s not looping from

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Is there a more efficient way to write this so it’s not looping from 1 to n (which hangs on n == 2**32):

def ns_num(n, seed, modulo, incrementor):
    assert n < modulo

    current = seed # some start value
    for i in xrange(1, n):
        current = (current + incrementor) % modulo

    return current

print ns_num(5, 3250, 87178291199, 17180131327)
print ns_num(2**32, 3250, 87178291199, 17180131327)
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  1. Editorial Team
    Editorial Team
    2026-05-23T05:34:00+00:00Added an answer on May 23, 2026 at 5:34 am

    That’s the same as

    return (seed + (n - 1) * incrementor) % modulo
    

    (Are you sure you want n - 1? That’s what you current code does.)

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