結果
問題 |
No.1931 Fraction 2
|
ユーザー |
![]() |
提出日時 | 2025-04-15 21:23:23 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,113 bytes |
コンパイル時間 | 270 ms |
コンパイル使用メモリ | 81,704 KB |
実行使用メモリ | 75,524 KB |
最終ジャッジ日時 | 2025-04-15 21:29:26 |
合計ジャッジ時間 | 5,751 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 3 WA * 7 TLE * 1 -- * 25 |
ソースコード
import sys import math MOD = 998244353 def main(): sys.setrecursionlimit(1 << 25) N = int(sys.stdin.readline()) A = [] B = [] max_b = 0 for _ in range(N): a, b = map(int, sys.stdin.readline().split()) A.append(a) B.append(b) if b > max_b: max_b = b # Precompute smallest prime factors up to max_b SPF = [0] * (max_b + 1) for i in range(2, max_b + 1): if SPF[i] == 0: SPF[i] = i for j in range(i*i, max_b+1, i): if SPF[j] == 0: SPF[j] = i # Factorize each B_i and track max exponents from collections import defaultdict prime_max = defaultdict(int) factorizations = [] for b in B: factors = {} if b == 1: factorizations.append(factors) continue x = b while x != 1: p = SPF[x] cnt = 0 while x % p == 0: cnt += 1 x //= p factors[p] = cnt if cnt > prime_max[p]: prime_max[p] = cnt factorizations.append(factors) # Compute LCM_mod LCM_mod = 1 lcm_primes = list(prime_max.keys()) for p in lcm_primes: exp = prime_max[p] LCM_mod = LCM_mod * pow(p, exp, MOD) % MOD # Compute sum_num_mod sum_num_mod = 0 for i in range(N): a = A[i] b_factors = factorizations[i] m_i_mod = 1 for p in lcm_primes: exp = prime_max[p] if p in b_factors: e_p = b_factors[p] current_exp = exp - e_p else: current_exp = exp m_i_mod = m_i_mod * pow(p, current_exp, MOD) % MOD term = a * m_i_mod % MOD sum_num_mod = (sum_num_mod + term) % MOD # Compute GCD of sum_num_mod and LCM_mod g = math.gcd(sum_num_mod, LCM_mod) # Compute inverse of g modulo MOD inv_g = pow(g, MOD-2, MOD) c = sum_num_mod * inv_g % MOD d = LCM_mod * inv_g % MOD print(c, d) if __name__ == '__main__': main()