結果

問題 No.3038 シャッフルの再現
ユーザー gew1fw
提出日時 2025-06-12 15:41:14
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,594 bytes
コンパイル時間 340 ms
コンパイル使用メモリ 82,380 KB
実行使用メモリ 67,784 KB
最終ジャッジ日時 2025-06-12 15:41:17
合計ジャッジ時間 2,679 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
（要ログイン）
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
import math
from collections import defaultdict

MOD = 10**9 + 7

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

def get_divisors(n):
    factors = factorize(n)
    divisors = [1]
    for p, exp in factors.items():
        current_length = len(divisors)
        p_power = 1
        for _ in range(exp):
            p_power *= p
            for d in divisors[:current_length]:
                divisors.append(d * p_power)
    divisors = sorted(divisors)
    return divisors

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

def compute_pisano_period(p):
    if p == 2:
        return 3
    if p == 5:
        return 20
    mod5 = p % 5
    if mod5 in (1, 4):
        number = p - 1
    else:
        number = 2 * (p + 1)
    divisors = get_divisors(number)
    for d in divisors:
        fd, fd1 = fast_doubling(d, p)
        if fd == 0 and fd1 == 1 % p:
            return d
    return None  # Should not reach here

def main():
    input = sys.stdin.read().split()
    ptr = 0
    n = int(input[ptr])
    ptr += 1
    global_factors = defaultdict(int)
    for _ in range(n):
        p = int(input[ptr])
        k = int(input[ptr + 1])
        ptr += 2
        if p == 2:
            if k == 1:
                pi_pk = 3
            else:
                pi_pk = 3 * (2 ** (k - 1))
            factors = factorize(pi_pk)
        elif p == 5:
            pi_pk = 4 * (5 ** k)
            factors = factorize(pi_pk)
        else:
            pi_p = compute_pisano_period(p)
            mod_p_sq = p * p
            fd_pi_p = fast_doubling(pi_p, mod_p_sq)[0]
            if fd_pi_p % mod_p_sq == 0:
                m = k
            else:
                m = k - 1
            factors = factorize(pi_p)
            factors[p] += m
        for prime, exp in factors.items():
            if global_factors[prime] < exp:
                global_factors[prime] = exp
    result = 1
    for prime, exp in global_factors.items():
        result = (result * pow(prime, exp, MOD)) % MOD
    print(result)

if __name__ == "__main__":
    main()
0