結果

問題 No.3038 シャッフルの再現
コンテスト
ユーザー gew1fw
提出日時 2025-06-12 18:50:40
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 3,486 bytes
コンパイル時間 220 ms
コンパイル使用メモリ 82,396 KB
実行使用メモリ 68,140 KB
最終ジャッジ日時 2025-06-12 18:50:49
合計ジャッジ時間 2,547 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 sys
from math import gcd

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(factors):
    divisors = [1]
    for p in sorted(factors.keys()):
        exp = factors[p]
        current_powers = [p**e for e in range(1, exp+1)]
        new_divisors = []
        for d in divisors:
            for power in current_powers:
                new_divisors.append(d * power)
        merged = []
        i = j = 0
        len_div = len(divisors)
        len_new = len(new_divisors)
        while i < len_div and j < len_new:
            if divisors[i] < new_divisors[j]:
                merged.append(divisors[i])
                i += 1
            else:
                merged.append(new_divisors[j])
                j += 1
        merged += divisors[i:]
        merged += new_divisors[j:]
        divisors = merged
    return divisors

def fast_doubling_pair(n, mod):
    if n == 0:
        return (0, 1)
    a, b = 0, 1
    s = bin(n)[2:]
    for bit in s:
        c = a * ((2 * b - a) % mod)
        d = (a * a + b * b) % mod
        if bit == '1':
            a, b = d, (c + d) % mod
        else:
            a, b = c, d
    return (a, b)

def compute_tau(p):
    if p == 2:
        return 3
    if p == 5:
        return 20
    mod5 = p % 5
    if mod5 in (1, 4):
        m = p - 1
    else:
        m = 2 * (p + 1)
    m_factors = factorize(m)
    divisors = generate_divisors(m_factors)
    for d in divisors:
        fd, fd1 = fast_doubling_pair(d, p)
        if fd == 0 and fd1 == 1:
            return d
    return m

def get_tau_factors(tau, m_factors):
    tau_factors = {}
    for p in m_factors:
        e = 0
        while tau % p == 0:
            e += 1
            tau = tau // p
        if e > 0:
            tau_factors[p] = e
    if tau != 1:
        pass
    return tau_factors

def main():
    input = sys.stdin.read().split()
    ptr = 0
    n = int(input[ptr])
    ptr += 1
    lcm_factors = {}
    for _ in range(n):
        p = int(input[ptr])
        k = int(input[ptr+1])
        ptr +=2
        if p == 2:
            if k == 1:
                current_factors = {3:1}
            elif k == 2:
                current_factors = {2:1, 3:1}
            else:
                current_factors = {2: (k-1), 3:1}
        elif p ==5:
            current_factors = {2:2, 5:k}
        else:
            mod5 = p %5
            if mod5 in (1,4):
                m_base = p-1
            else:
                m_base = 2*(p+1)
            m_factors = factorize(m_base)
            tau_p = compute_tau(p)
            tau_factors = get_tau_factors(tau_p, m_factors)
            current_factors = tau_factors.copy()
            if p in current_factors:
                current_factors[p] += (k-1)
            else:
                current_factors[p] = (k-1)
        for prime, exp in current_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 lcm_factors.items():
        result = (result * pow(prime, exp, MOD)) % MOD
    print(result)

if __name__ == "__main__":
    main()
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