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/*
* Suppose we have n = q * d, all integers. We know n and d, and want q = n / d.
*
* For any k, we have (here, all division is exact; not C-style rounding):
* floor(ceil(2^k / d) * n / 2^k) = floor((2^k + r) / d * n / 2^k), where
* r = (-2^k) mod d.
*
* Expanding this out:
* ... = floor(2^k / d * n / 2^k + r / d * n / 2^k)
* = floor(n / d + (r / d) * (n / 2^k)).
*
* The fractional part of n / d is 0 (because of the assumption that d divides n
* exactly), so we have:
* ... = n / d + floor((r / d) * (n / 2^k))
*
* So that our initial expression is equal to the quantity we seek, so long as
* (r / d) * (n / 2^k) < 1.
*
* r is a remainder mod d, so r < d and r / d < 1 always. We can make
* n / 2 ^ k < 1 by setting k = 32. This gets us a value of magic that works.
*/
void