結果
問題 |
No.2954 Calculation of Exponentiation
|
ユーザー |
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提出日時 | 2025-03-31 17:51:09 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,834 bytes |
コンパイル時間 | 260 ms |
コンパイル使用メモリ | 82,692 KB |
実行使用メモリ | 58,588 KB |
最終ジャッジ日時 | 2025-03-31 17:52:04 |
合計ジャッジ時間 | 2,199 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 25 WA * 3 |
ソースコード
import mathdef gcd(a, b):while b:a, b = b, a % breturn adef parse_fraction(s):if '.' in s:left, right = s.split('.')else:left = sright = '0000'right = right.ljust(4, '0')[:4]numerator = int(left) * 10000 + int(right)denominator = 10000g = gcd(numerator, denominator)return numerator // g, denominator // gdef parse_fraction_with_sign(s):sign = 1if s.startswith('-'):sign = -1s = s[1:]if '.' in s:left, right = s.split('.')else:left = sright = '0000'right = right.ljust(4, '0')[:4]numerator = sign * (int(left) * 10000 + int(right))denominator = 10000g = gcd(abs(numerator), denominator)return numerator // g, denominator // gdef factor(n):factors = {}if n == 0:return factorswhile n % 2 == 0:factors[2] = factors.get(2, 0) + 1n = n // 2i = 3max_i = math.isqrt(n) + 1while i <= max_i and n > 1:while n % i == 0:factors[i] = factors.get(i, 0) + 1n = n // imax_i = math.isqrt(n) + 1i += 2if n > 1:factors[n] = 1return factorsA, B = input().split()a_num, a_den = parse_fraction(A)b_num, b_den = parse_fraction_with_sign(B)if b_num == 0:print("Yes")else:p = a_numq = a_denc = b_numd = b_denabs_c = abs(c)p_factors = factor(p)q_factors = factor(q)valid = Truefor prime, exp in p_factors.items():if (exp * abs_c) % d != 0:valid = Falsebreakif valid:for prime, exp in q_factors.items():if (exp * abs_c) % d != 0:valid = Falsebreakprint("Yes" if valid else "No")