結果
| 問題 |
No.2210 equence Squence Seuence
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-31 18:00:05 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,758 bytes |
| コンパイル時間 | 219 ms |
| コンパイル使用メモリ | 82,628 KB |
| 実行使用メモリ | 94,844 KB |
| 最終ジャッジ日時 | 2025-03-31 18:01:02 |
| 合計ジャッジ時間 | 4,765 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | TLE * 1 -- * 24 |
ソースコード
import sys
from functools import cmp_to_key
def main():
N, K = map(int, sys.stdin.readline().split())
A = list(map(int, sys.stdin.readline().split()))
# This function compares two indices by their corresponding X sequences
def compare(i, j):
if i == j:
return 0
# Iterate through elements of X_i and X_j to find the first difference
ii = 0 # current position in X_i and X_j
while True:
# Skip the current index for each sequence
xi = ii if ii < i else ii + 1
xj = ii if ii < j else ii + 1
elem_i = A[xi] if xi < N else None
elem_j = A[xj] if xj < N else None
# Compare elements at current position
if elem_i < elem_j:
return -1
elif elem_i > elem_j:
return 1
else:
ii += 1
# Break if both sequences have been exhausted
if (xi >= N-1 and xj >= N-1) or ii >= (N-1):
break
# All elements are the same up to the length of the shorter sequence
# So, the one with smaller original index comes first
return -1 if i < j else 1
# Create a list of indices from 0 to N-1 (original indices)
indices = list(range(N))
# Sort the indices based on the compare function
sorted_indices = sorted(indices, key=cmp_to_key(compare))
# Get the K-th smallest index (converting K from 1-based to 0-based)
selected = sorted_indices[K-1]
# Generate the output sequence by skipping the selected index
output = [str(A[i]) for i in range(N) if i != selected]
print(' '.join(output))
if __name__ == "__main__":
main()
lam6er