結果
問題 |
No.3038 シャッフルの再現
|
ユーザー |
![]() |
提出日時 | 2025-06-12 18:49:00 |
言語 | PyPy3 (7.3.15) |
結果 |
RE
|
実行時間 | - |
コード長 | 1,977 bytes |
コンパイル時間 | 166 ms |
コンパイル使用メモリ | 82,104 KB |
実行使用メモリ | 67,676 KB |
最終ジャッジ日時 | 2025-06-12 18:49:18 |
合計ジャッジ時間 | 2,582 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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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 from math import gcd from functools import reduce MOD = 10**9 + 7 def factorize(m): factors = [] i = 2 while i * i <= m: if m % i == 0: cnt = 0 while m % i == 0: cnt += 1 m = m // i factors.append((i, cnt)) i += 1 if m > 1: factors.append((m, 1)) return factors def generate_divisors(factors): divisors = [1] for (p, exp) in factors: temp = [] for d in divisors: current = d for _ in range(exp + 1): temp.append(current) current *= p divisors = temp divisors = sorted(divisors) return divisors def fast_doubling(n, mod): if n == 0: return (0, 1) a, b = fast_doubling(n >> 1, mod) c = (a * (2 * b - a)) % mod d = (a * a + b * b) % mod if n & 1: return (d, (c + d) % mod) else: return (c, d) def get_pisano_period(p): if p == 5: return 20 mod5 = p % 5 if mod5 in (1, 4): m = p - 1 else: m = 2 * (p + 1) factors = factorize(m) divisors = generate_divisors(factors) for d in divisors: if d == 0: continue fn, fn1 = fast_doubling(d, p) if fn % p == 0 and fn1 % p == 1: return d return m # Fallback, though theoretically should not reach here def lcm(a, b): return a * b // gcd(a, b) def main(): input = sys.stdin.read().split() idx = 0 N = int(input[idx]) idx +=1 periods = [] for _ in range(N): p = int(input[idx]) k = int(input[idx+1]) idx +=2 base_period = get_pisano_period(p) period = base_period * (p ** (k-1)) periods.append(period) # Compute LCM of all periods if not periods: print(0) return total_lcm = reduce(lcm, periods) print(total_lcm % MOD) if __name__ == '__main__': main()