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 16:22:11
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,694 bytes
コンパイル時間 216 ms
コンパイル使用メモリ 81,708 KB
実行使用メモリ 67,244 KB
最終ジャッジ日時 2025-04-16 16:23:24
合計ジャッジ時間 2,248 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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 # 

MOD = 10**9 + 7

def factorize(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_sorted(factors):
    divisors = [1]
    for p in sorted(factors.keys()):
        exp = factors[p]
        current_powers = [p**e for e in range(exp + 1)]
        new_divisors = []
        for d in divisors:
            for power in current_powers:
                new_divisors.append(d * power)
        divisors = sorted(list(set(new_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 compute_pisano_prime(p):
    if p == 2:
        return 3
    if p == 5:
        return 20
    legendre = pow(5, (p - 1) // 2, p)
    if legendre == 1 or legendre == 0:
        candidate = p - 1
    else:
        candidate = 2 * (p + 1)
    factors = factorize(candidate)
    divisors = generate_divisors_sorted(factors)
    for d in divisors:
        if d == 0:
            continue
        f_d, f_d_plus_1 = fast_doubling(d, p)
        if f_d % p == 0 and f_d_plus_1 % p == 1:
            return d
    return candidate

def get_pisano_power_factors(p, k):
    if p == 2:
        if k == 1:
            return {3: 1}
        elif k == 2:
            return {2: 1, 3: 1}
        else:
            return {2: (k - 1), 3: 1}
    elif p == 5:
        return {2: 2, 5: k}
    else:
        pisano_p = compute_pisano_prime(p)
        factors = factorize(pisano_p)
        factors[p] = factors.get(p, 0) + (k - 1)
        return factors

def main():
    import sys
    input = sys.stdin.read().split()
    idx = 0
    while idx < len(input):
        N = int(input[idx])
        idx += 1
        lcm_factors = {}
        for _ in range(N):
            p = int(input[idx])
            k = int(input[idx + 1])
            idx += 2
            factors = get_pisano_power_factors(p, k)
            for prime, exp in factors.items():
                if prime in lcm_factors:
                    if exp > lcm_factors[prime]:
                        lcm_factors[prime] = exp
                else:
                    lcm_factors[prime] = exp
        result = 1
        for prime, exp in sorted(lcm_factors.items()):
            result = (result * pow(prime, exp, MOD)) % MOD
        print(result)

if __name__ == "__main__":
    main()
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