結果

問題 No.3038 シャッフルの再現
ユーザー lam6er
提出日時 2025-03-20 20:37:29
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,610 bytes
コンパイル時間 145 ms
コンパイル使用メモリ 82,420 KB
実行使用メモリ 67,872 KB
最終ジャッジ日時 2025-03-20 20:38:19
合計ジャッジ時間 2,434 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 math

MOD = 10**9 + 7

def trial_division(n):
    factors = {}
    while n % 2 == 0:
        factors[2] = factors.get(2, 0) + 1
        n = n // 2
    i = 3
    max_factor = int(math.isqrt(n)) + 1
    while i <= max_factor and n > 1:
        while n % i == 0:
            factors[i] = factors.get(i, 0) + 1
            n = n // i
            max_factor = int(math.isqrt(n)) + 1
        i += 2
    if n > 1:
        factors[n] = 1
    return factors

def generate_divisors(factors):
    divisors = [1]
    for prime in factors:
        exponents = [prime**i for i in range(1, factors[prime] + 1)]
        new_divisors = []
        for d in divisors:
            for e in exponents:
                new_divisors.append(d * e)
        divisors += new_divisors
    divisors = list(set(divisors))
    divisors.sort()
    return divisors

def fib_pair(n, mod):
    if n == 0:
        return (0, 1)
    a, b = fib_pair(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 compute_pisano_factors(p):
    if p == 2:
        return {2: 0, 3: 1}, 3
    elif p == 5:
        return {2: 2, 5: 1}, 20
    if p % 5 in (1, 4):
        candidate_num = p - 1
    else:
        candidate_num = 2 * (p + 1)
    factors_candidate = trial_division(candidate_num)
    divisors = generate_divisors(factors_candidate)
    d = None
    for current_d in divisors:
        fn, fn1 = fib_pair(current_d, p)
        if fn == 0 and fn1 == 1:
            d = current_d
            break
    factors_d = {}
    for prime in factors_candidate:
        count = 0
        temp_d = d
        while temp_d % prime == 0:
            count += 1
            temp_d = temp_d // prime
        if count > 0:
            factors_d[prime] = count
    return factors_d, d

def main():
    n = int(input())
    prime_powers = [tuple(map(int, input().split())) for _ in range(n)]
    lcm_factors = {}
    for p, k in prime_powers:
        if p == 2:
            period_factors = {2: k-1, 3: 1}
        elif p == 5:
            period_factors = {2: 2, 5: k}
        else:
            factors_d, _ = compute_pisano_factors(p)
            period_factors = factors_d.copy()
            period_factors[p] = period_factors.get(p, 0) + (k - 1)
        for prime, exp in period_factors.items():
            if exp > lcm_factors.get(prime, 0):
                lcm_factors[prime] = exp
    result = 1
    for prime, exp in lcm_factors.items():
        result = (result * pow(prime, exp, MOD)) % MOD
    print(result)

if __name__ == "__main__":
    main()
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