結果

問題 No.1036 Make One With GCD 2
ユーザー 草苺奶昔
提出日時 2023-03-10 18:28:13
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 767 ms / 2,000 ms
コード長 2,283 bytes
コンパイル時間 213 ms
コンパイル使用メモリ 82,032 KB
実行使用メモリ 253,808 KB
最終ジャッジ日時 2024-09-18 03:20:40
合計ジャッジ時間 19,543 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 4
other AC * 41
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

from math import gcd
from typing import Callable, Generic, List, TypeVar
E = TypeVar("E")
class SlidingWindowAggregation(Generic[E]):
"""SlidingWindowAggregation
Api:
1. append value to tail,O(1).
2. pop value from head,O(1).
3. query aggregated value in window,O(1).
"""
__slots__ = ["_stack0", "_stack1", "_stack2", "_stack3", "_e0", "_e1", "_size", "_op", "_e"]
def __init__(self, e: Callable[[], E], op: Callable[[E, E], E]):
"""
Args:
e: unit element
op: merge function
"""
self._stack0 = []
self._stack1 = []
self._stack2 = []
self._stack3 = []
self._e = e
self._e0 = e()
self._e1 = e()
self._size = 0
self._op = op
def append(self, value: E) -> None:
if not self._stack0:
self._push0(value)
self._transfer()
else:
self._push1(value)
self._size += 1
def popleft(self) -> None:
if not self._size:
return
if not self._stack0:
self._transfer()
self._stack0.pop()
self._stack2.pop()
self._e0 = self._stack2[-1] if self._stack2 else self._e()
self._size -= 1
def query(self) -> E:
return self._op(self._e0, self._e1)
def _push0(self, value):
self._stack0.append(value)
self._e0 = self._op(value, self._e0)
self._stack2.append(self._e0)
def _push1(self, value):
self._stack1.append(value)
self._e1 = self._op(self._e1, value)
self._stack3.append(self._e1)
def _transfer(self):
while self._stack1:
self._push0(self._stack1.pop())
while self._stack3:
self._stack3.pop()
self._e1 = self._e()
@property
def size(self):
return self._size
if __name__ == "__main__":
n = int(input())
nums = list(map(int, input().split()))
S = SlidingWindowAggregation(lambda: 0, gcd)
res = 0
right = 0
for left in range(n):
right = max(right, left)
while right < n and gcd(S.query(), nums[right]) != 1:
S.append(nums[right])
right += 1
res += n - right
S.popleft()
print(res)
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