python 題目求指點

45 回覆
2 Like 4 Dislike
2019-02-21 18:14:14



自己個code


我嘗試去run 佢但係有several case 係runtime error 而我又睇唔到個d case 有無巴打可以指點下我話比我聽個code ge盲眼位係邊到 thank you
2019-02-21 18:15:47
手機影mon
2019-02-21 18:25:08
2019-02-21 19:38:21
發現樓主
https://stackoverflow.com/questions/54789625/pillar-jumpers-in-python
2019-02-21 19:50:44
一撚定係啦 仲識上lihkg
2019-02-21 19:51:36
有無人幫手
2019-02-21 19:56:04
S 係 strength
即係佢可以由i跳去i+S咁?
2019-02-21 19:57:14
係咪啲loop loop好撚多次呀？
2019-02-21 20:00:33
係 我發現我漏左呢樣野 但我唔知點樣可以add 落個function
2019-02-21 20:01:56
呢個都係其中一個原因 咁我可以點樣改善呢個問題
2019-02-21 20:10:30
S = N-1 no matter what J is.
利伸:9up
2019-02-21 20:39:53
你唔講你段code 既logic 叫人點幫你呀
2019-02-21 20:43:58
手機影 又唔講你段code 既logic 點 run , 好難幫到你
唯有幫你google 下

https://stackoverflow.com/questions/54789625/pillar-jumpers-in-python
2019-02-21 20:51:50
用S 來做binary search

Start = max difference between each pillar

End= Number of pillar/J
2019-02-21 23:20:26
但係要搵minimum strength ge 話 用binary search 唔會好唔方便咩
2019-02-21 23:25:16
點解樓主唔用pycharm
靚好多function好好多
2019-02-21 23:27:10
Dynamic programming 會唔會好啲
2019-02-21 23:35:40
唔會，binary search 每次check canJump(S) 好簡單咋喎
2019-02-21 23:46:42
我而岩岩起步學programming 可唔可以講多d thx
2019-02-21 23:47:50
PyCharm係ide嚟
如果你係學生既話條license係免費
2019-02-21 23:49:06
我試下先
同埋想問下點樣可以將 max_jump = strength + current_position 然後用max_jump 取代function 入面個個for loop 個上限 from number_of_pillars to max_jump
2019-02-21 23:50:09
個陣我要install 個platform 然後個frd 叫我用spyder 唔好煩 而家down 黎試下撚下先
2019-02-21 23:51:00
你個fd真係夠晒老友
2019-02-21 23:51:30
而家去左exchange 我都係識左佢幾日
2019-02-21 23:53:16
去jetbrains註冊個Student ac先
然後verify個學生email address就會有free license比你用
裝咗之後paste你d code落pycharm睇下佢會唔會出咩error msg或者suggestion
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