→ rockieloser: log運算而已 01/22 21:08
→ rockieloser: log(n/2^i) = log(n) - log(2^i) (log2^i = i) 01/22 21:18
您好,log(2^i)=i這部分請問是2為底嗎?
※ 編輯: OwTaingJune (36.225.121.51), 01/22/2019 21:22:33
※ 編輯: OwTaingJune (36.225.121.51), 01/22/2019 21:23:25
→ rockieloser: 是 01/22 21:27
好的,謝謝您!
※ 編輯: OwTaingJune (36.225.121.51), 01/22/2019 22:05:17