結果

問題 No.278 連続する整数の和(2)
ユーザー ThetaTheta
提出日時 2022-11-25 17:17:52
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 241 ms / 2,000 ms
コード長 1,641 bytes
コンパイル時間 154 ms
コンパイル使用メモリ 12,800 KB
実行使用メモリ 11,136 KB
最終ジャッジ日時 2024-10-02 00:16:03
合計ジャッジ時間 1,983 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
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
other AC * 18
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 itertools import count
from math import prod
def calc_positive_divisors(num: int) -> list[int]:
small_divisors = []
large_divisors = []
for n in range(1, floor(sqrt(num))+1):
if num % n == 0:
small_divisors.append(n)
large_divisors.append(num//n)
if small_divisors[-1] == large_divisors[-1]:
large_divisors.pop()
divisors = small_divisors + reversed(large_divisors)
return divisors
def calc_prime_factorize(num: int) -> dict[int, int]:
if num < 2:
raise ValueError
divisors = {}
for divisor in count(2):
if divisor ** 2 > num:
if num != 1:
divisors[num] = 1
break
while num % divisor == 0 and num != 1:
try:
divisors[divisor] += 1
except KeyError:
divisors[divisor] = 1
num //= divisor
return divisors
def gcd(*numbers: int) -> int:
if len(numbers) == 1:
return numbers[0]
if len(numbers) == 2:
a, b = numbers
if a < b:
a, b = b, a
while True:
if a % b == 0:
return b
a, b = b, a % b
first_gcd = gcd(*numbers[:2])
return gcd(first_gcd, *numbers[2:])
def main():
N = int(input())
if N == 1:
print(1)
return
max_X = gcd(N, (N-1)*N//2)
if max_X == 1:
print(1)
return
factors = calc_prime_factorize(max_X)
print(prod(sum(factor ** pow_ for pow_ in range(number+1))
for factor, number in factors.items()))
if __name__ == "__main__":
main()
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