結果

問題 No.3038 シャッフルの再現
ユーザー lam6er
提出日時 2025-04-16 00:15:48
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,396 bytes
コンパイル時間 166 ms
コンパイル使用メモリ 81,824 KB
実行使用メモリ 67,236 KB
最終ジャッジ日時 2025-04-16 00:17:35
合計ジャッジ時間 2,028 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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
sample RE * 1
other RE * 21
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 #

import sys
from math import gcd
from functools import reduce

MOD = 10**9 + 7

def factor(n):
    factors = {}
    while n % 2 == 0:
        factors[2] = factors.get(2, 0) + 1
        n = n // 2
    i = 3
    while i * i <= n:
        while n % i == 0:
            factors[i] = factors.get(i, 0) + 1
            n = n // i
        i += 2
    if n > 1:
        factors[n] = 1
    return factors

def generate_divisors(factors):
    divisors = [1]
    for p, exp in factors.items():
        current_divisors = []
        for d in divisors:
            current = 1
            for _ in range(exp + 1):
                current_divisors.append(d * current)
                current *= p
        divisors = current_divisors
    divisors = list(set(divisors))
    divisors.sort()
    return divisors

def fib_pair(n, mod):
    if mod == 1:
        return (0, 0)
    def fib(n_val):
        if n_val == 0:
            return (0, 1)
        a, b = fib(n_val >> 1)
        c = (a * (2 * b - a)) % mod
        d = (a * a + b * b) % mod
        if n_val & 1:
            return (d, (c + d) % mod)
        else:
            return (c, d)
    return fib(n)

def compute_pisano_period(p):
    if p == 2:
        return 3
    if p == 5:
        return 20
    legendre = pow(5, (p - 1) // 2, p)
    if legendre == 1:
        m = p - 1
    else:
        m = 2 * (p + 1)
    factors = factor(m)
    divisors = generate_divisors(factors)
    for d in divisors:
        if d == 0:
            continue
        fib_d, fib_d_plus_1 = fib_pair(d, p)
        if fib_d % p == 0 and fib_d_plus_1 % p == 1:
            return d
    return m

def compute_period_prime_power(p, k):
    if p == 2:
        if k == 1:
            return 3
        elif k == 2:
            return 6
        else:
            return 3 * (2 ** (k - 1))
    elif p == 5:
        return 20 * (5 ** (k - 1))
    else:
        pisano_p = compute_pisano_period(p)
        return pisano_p * (p ** (k - 1))

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
        period = compute_period_prime_power(p, k)
        periods.append(period)
    def lcm(a, b):
        return a * b // gcd(a, b)
    result = reduce(lcm, periods, 1)
    print(result % MOD)

if __name__ == '__main__':
    main()
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