#868130 (PyPy3) No.2210 equence Squence Seuence

• 問題: No.2210 equence Squence Seuence
• ユーザー: kuro_B
• 提出日時: 2023-05-12 15:33:03
• 言語: PyPy3 (7.3.15)
• 結果: AC
• 実行時間: 230 ms / 2,000 ms
• コード長: 1,739 bytes
• コンパイル時間: 430 ms
• コンパイル使用メモリ: 82,240 KB
• 実行使用メモリ: 122,800 KB
• 最終ジャッジ日時: 2024-11-28 12:52:32
• 合計ジャッジ時間: 7,971 ms
• ジャッジサーバーID （参考情報）: judge1 / judge4

ファイルパターン	結果
sample	AC * 3
other	AC * 25

ソースコード

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###スニペット始まり###
import sys, re
from copy import copy, deepcopy
from math import ceil, floor, sqrt,factorial, gcd, pi, degrees, radians, sin, asin, cos, acos, tan, atan2
from statistics import mean, median
from collections import Counter, deque, defaultdict
from heapq import heapify, heappop, heappush
from itertools import permutations, accumulate, product, combinations, combinations_with_replacement
from bisect import bisect, bisect_left, bisect_right
from functools import reduce, lru_cache
from string import ascii_uppercase, ascii_lowercase
from decimal import Decimal, ROUND_HALF_UP #四捨五入用
def input(): return sys.stdin.readline().rstrip('\n')
#easy-testのpypyでは再帰数を下げる。
if __file__=='prog.py':
    sys.setrecursionlimit(10**5)
else:
    sys.setrecursionlimit(10**6)
def lcm(a, b): return a * b // gcd(a, b)
#3つ以上の最大公約数/最小公倍数。Nを要素数、Mを数値の大きさとして、O(NlogM)
def gcd_v2(l: list): return reduce(gcd, l)
def lcm_v2(l: list): return reduce(lcm, l)
#nPk
def nPk(n, k): return factorial(n) // factorial(n - k)
#逆元
def modinv(a, mod=10**9+7): return pow(a, mod-2, mod)
INF = float('inf')
MOD = 10 ** 9 + 7
###スニペット終わり###
N, K=map(int, input().split())
A=list(map(int, input().split()))
A.append(-999)#番兵
res=[-1]*N
L=0
R=N-1
stock=[]
for i in range(N):
    stock.append(i)
    if A[i]>A[i+1]:
        while stock:
            res[L]=stock[-1]
            L+=1
            stock.pop()
    elif A[i]<A[i+1]:
        while stock:
            res[R]=stock[-1]
            R-=1
            stock.pop()
ans_idx=res[K-1]
ans=[]
for i,a in enumerate(A[:N]):
    if i!=ans_idx:
        ans.append(a)
print(*ans)
 
 
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