這是
Note that the plot that was previously here was wrong! The code and data were not though. The code has been updated to include @MarkRansom's solution, since it turns out to be the optimal solution for very large numbers (over 4k-bit numbers).
uj5u.com熱心網友回復：
while (num & 0xffffffff) == 0:
num >>= 32
if (num & 0xffff) == 0:
num >>= 16
if (num & 0xff) == 0:
num >>= 8
if (num & 0xf) == 0:
num >>= 4
if (num & 0x3) == 0:
num >>= 2
if (num & 0x1) == 0:
num >>= 1
這里的想法是執行盡可能少的班次。初始while回圈處理超過 32 位長的數字，我認為這不太可能，但必須提供它以確保完整性。之后，每條陳述句移動一半的位；如果你不能移動 16，那么你最多可以移動 15，即 (8 4 2 1)。if這 5 個陳述涵蓋了所有可能的情況。
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