結果
問題 |
No.3038 シャッフルの再現
|
ユーザー |
![]() |
提出日時 | 2025-03-31 17:33:32 |
言語 | PyPy3 (7.3.15) |
結果 |
RE
|
実行時間 | - |
コード長 | 2,128 bytes |
コンパイル時間 | 206 ms |
コンパイル使用メモリ | 82,400 KB |
実行使用メモリ | 67,760 KB |
最終ジャッジ日時 | 2025-03-31 17:34:12 |
合計ジャッジ時間 | 2,374 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | RE * 1 |
other | RE * 21 |
ソースコード
import sys MOD = 10**9 + 7 def fast_doubling(n, p): if n == 0: return (0, 1) a, b = fast_doubling(n >> 1, p) c = (a * ((2 * b - a) % p)) % p d = (a * a + b * b) % p if n & 1: return (d, (c + d) % p) else: return (c, d) def check_pisano(d, p): if d == 0: return False a, b = fast_doubling(d, p) return (a % p == 0) and (b % p == 1) def factor(n): factors = {} while n % 2 == 0: factors[2] = factors.get(2, 0) + 1 n //= 2 i = 3 while i * i <= n: while n % i == 0: factors[i] = factors.get(i, 0) + 1 n //= i i += 2 if n > 1: factors[n] = 1 return factors def generate_divisors(factors): divisors = [1] for p, exp in factors.items(): temp = [] for d in divisors: current = 1 for _ in range(exp + 1): temp.append(d * current) current *= p divisors = temp return divisors def compute_period_pisano(p): if p == 2: return 3 if p == 5: return 20 legendre = pow(5, (p - 1) // 2, p) if legendre == 1 or legendre == 0: target = p - 1 else: target = 2 * (p + 1) factors = factor(target) divisors = generate_divisors(factors) divisors = sorted(divisors) for d in divisors: if d == 0: continue if check_pisano(d, p): return d return target def lcm(a, b): from math import gcd return a * b // gcd(a, b) def main(): input = sys.stdin.read().split() idx = 0 N = int(input[idx]) idx += 1 current_lcm = 1 for _ in range(N): p = int(input[idx]) k = int(input[idx + 1]) idx += 2 if p == 2: pe_pisano = 3 * (2 ** (k - 1)) elif p == 5: pe_pisano = 20 * (5 ** (k - 1)) else: pi_p = compute_period_pisano(p) pe_pisano = pi_p * (p ** (k - 1)) current_lcm = lcm(current_lcm, pe_pisano) print(current_lcm % MOD) if __name__ == '__main__': main()