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1. Commits on Feb 25, 2020

   • Merge pull request #1 from blavejr/patch-2

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2. Commits on Oct 31, 2019

   • Update frog_jmp.py

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3. Commits on Jun 23, 2019

   • 應該為求N與M的最大公因數X後，回傳N/X

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   • 使用N+1陣列，這樣較好理解

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4. Commits on Feb 28, 2019

   • 移除.gitignore

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   • 查詢官方解答後，複雜度為O(N * max(abs(A))**2)，可以拿到100%。

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   • 查詢官方解答，複雜度為O(N**2*max(abs(A)))，可以拿到63%。

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   • 查詢網路作法後，得知需使用Dynamic programming依序計算骰子丟出結果，完成測驗。

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5. Commits on Feb 27, 2019

   • 建立檔案

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   • 查詢網路作法後，得知原來按照順序迴圈計算即可，完成測驗。

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   • 使用deque與反向排序，拿到25%

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   • 查詢網路作法後，完成測驗。

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   • 無法理解為何codility回應testcase 2的正確答案是3

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6. Commits on Feb 26, 2019

   • 建立檔案

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   • 查詢網路作法後，得知若先將A排序，則min abs sum必定在差距最小的連續兩個數列上，故可將複雜度降為O(N * log(N))，完成測驗。

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   • 依照題目使用雙迴圈，遇到performance issue，拿到36%。

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   • 依照官方教材，複雜度降為O(N**2)，完成測驗。

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   • 使用迴圈count，複雜度為O(N**3)，拿到63%。

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   • 查詢網路作法後，複雜度降為O(N)，完成測驗。

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   • 使用雙迴圈count，拿到70%。

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   • 複雜度為O(N) or O(N*log(N))，完成測驗。

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   • 建立檔案

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   • 查詢網路作法後，完成測驗，複雜度為O(N*log(N+M))。

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7. Commits on Feb 25, 2019

   • 按照題目要求，使用迴圈求最大值。拿到25%。

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   • 補上參考資料

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   • 改用binary search解決performance issue，完成測驗。

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   • 查詢網路作法後，了解題目意思，修正判斷式之後，拿到50%，但仍有performance issue，複雜度為O((N+M)*N)。

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   • 使用雙迴圈計算，拿到25%。

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8. Commits on Feb 24, 2019

   • 建立檔案

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   • 用 & + shift 作法取代 modulo，複雜度降為O(L)，拿到100%。

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   • 補上測驗結果

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   • 使用查表法，遇到performance issue，拿到87%。

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9. Commits on Feb 22, 2019

   • 參考網路解答後，以Dynamic programming方式解決，複雜度降為O(N*log(N))，達成測驗。

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   • 只拿到16%。

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   • 建立檔案

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