z*********8
发帖数: 2070 | 1 would anyone explain why the running time is O(L^2), assume the numbers u, v
both have L bits.
Each step either or both u and v will be reduced by 1 bit, so totally there
would be O(L) steps, right? | t****t
发帖数: 6806 | 2 通常GCD这样的算法是要考虑位数的,因为通常位数会非常大.
v
there
【在 z*********8 的大作中提到】
: would anyone explain why the running time is O(L^2), assume the numbers u, v
: both have L bits.
: Each step either or both u and v will be reduced by 1 bit, so totally there
: would be O(L) steps, right?
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