結果

問題 No.1077 Noelちゃんと星々4
ユーザー Nagisa
提出日時 2020-06-12 22:42:42
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 276 ms / 2,000 ms
コード長 1,576 bytes
コンパイル時間 408 ms
コンパイル使用メモリ 82,360 KB
実行使用メモリ 153,884 KB
最終ジャッジ日時 2024-06-24 05:27:18
合計ジャッジ時間 4,698 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 20
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

def getMinimumOps(ar):
    # Number of elements in the array
    n = len(ar)
    # Smallest element in the array
    small = min(ar)
    # Largest element in the array
    large = max(ar)
    """
        dp(i, j) represents the minimum number
        of operations needed to make the
        array[0 .. i] sorted in non-decreasing
        order given that ith element is j
    """
    dp = [[ 0 for i in range(large + 1)]
              for i in range(n)]
    # Fill the dp[]][ array for base cases
    for j in range(small, large + 1):
        dp[0][j] = abs(ar[0] - j)
    """
    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    """
    for i in range(1, n):
        minimum = 10**9
        for j in range(small, large + 1):
        # """
        #     /*
        #     If the ith element is j then we can have
        #     any value from small to j for the i-1 th
        #     element
        #     We choose the one that requires the
        #     minimum operations
        # """
            minimum = min(minimum, dp[i - 1][j])
            dp[i][j] = minimum + abs(ar[i] - j)
    """
    /*
        If we made the (n - 1)th element equal to j
        we required dp(n-1, j) operations
        We choose the minimum among all possible
        dp(n-1, j) where j goes from small to large
    */
    """
    ans = 10**9
    for j in range(small, large + 1):
        ans = min(ans, dp[n - 1][j])
    return ans
N = int(input())
A = list(map(int,input().split()))
print(getMinimumOps(A))
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