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 | 267 ms / 2,000 ms |
コード長 | 1,576 bytes |
コンパイル時間 | 559 ms |
コンパイル使用メモリ | 87,292 KB |
実行使用メモリ | 155,184 KB |
最終ジャッジ日時 | 2023-09-06 10:51:46 |
合計ジャッジ時間 | 4,665 ms |
ジャッジサーバーID (参考情報) |
judge11 / judge12 |
(要ログイン)
テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 60 ms
71,184 KB |
testcase_01 | AC | 59 ms
71,168 KB |
testcase_02 | AC | 220 ms
130,328 KB |
testcase_03 | AC | 267 ms
146,060 KB |
testcase_04 | AC | 130 ms
97,476 KB |
testcase_05 | AC | 114 ms
90,700 KB |
testcase_06 | AC | 144 ms
103,856 KB |
testcase_07 | AC | 166 ms
107,804 KB |
testcase_08 | AC | 135 ms
99,212 KB |
testcase_09 | AC | 128 ms
96,328 KB |
testcase_10 | AC | 244 ms
145,928 KB |
testcase_11 | AC | 184 ms
121,636 KB |
testcase_12 | AC | 173 ms
113,416 KB |
testcase_13 | AC | 117 ms
90,360 KB |
testcase_14 | AC | 240 ms
144,212 KB |
testcase_15 | AC | 167 ms
112,436 KB |
testcase_16 | AC | 138 ms
99,844 KB |
testcase_17 | AC | 177 ms
117,020 KB |
testcase_18 | AC | 244 ms
147,568 KB |
testcase_19 | AC | 109 ms
90,412 KB |
testcase_20 | AC | 63 ms
75,108 KB |
testcase_21 | AC | 264 ms
155,184 KB |
ソースコード
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))