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 16:21:30
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 3,513 bytes
コンパイル時間 196 ms
コンパイル使用メモリ 81,996 KB
実行使用メモリ 70,100 KB
最終ジャッジ日時 2025-04-16 16:22:46
合計ジャッジ時間 2,458 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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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
import random
from collections import defaultdict

MOD = 10**9 + 7

def is_prime(n):
    if n < 2:
        return False
    for p in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
        if n % p == 0:
            return n == p
    d = n - 1
    s = 0
    while d % 2 == 0:
        d //= 2
        s += 1
    for a in [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]:
        if a >= n:
            continue
        x = pow(a, d, n)
        if x == 1 or x == n - 1:
            continue
        for _ in range(s - 1):
            x = pow(x, 2, n)
            if x == n - 1:
                break
        else:
            return False
    return True

def pollards_rho(n):
    if n % 2 == 0:
        return 2
    if n % 3 == 0:
        return 3
    if n % 5 == 0:
        return 5
    while True:
        c = random.randint(1, n-1)
        f = lambda x: (pow(x, 2, n) + c) % n
        x, y, d = 2, 2, 1
        while d == 1:
            x = f(x)
            y = f(f(y))
            d = math.gcd(abs(x - y), n)
        if d != n:
            return d

def factor(n):
    factors = []
    def _factor(n):
        if n == 1:
            return
        if is_prime(n):
            factors.append(n)
            return
        d = pollards_rho(n)
        _factor(d)
        _factor(n // d)
    _factor(n)
    factors.sort()
    return factors

def factorize(n):
    if n == 0:
        return {}
    factors = factor(n)
    res = defaultdict(int)
    for p in factors:
        res[p] += 1
    return res

def generate_divisors(factors_dict):
    factors = sorted(factors_dict.items())
    divisors = [1]
    for (p, exp) in factors:
        temp = []
        for d in divisors:
            current = 1
            for _ in range(exp + 1):
                temp.append(d * current)
                current *= p
        divisors = temp
    return sorted(divisors)

def fib_pair(n, mod):
    if mod == 1:
        return (0, 0)
    def fast_doubling(n):
        if n == 0:
            return (0, 1)
        a, b = fast_doubling(n >> 1)
        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)
    return fast_doubling(n)

def compute_pisano_period(p):
    if p == 2:
        return 3
    if p == 5:
        return 20
    a = 5
    legendre = pow(a, (p - 1) // 2, p)
    if legendre == 1 or legendre == 0:
        m = p - 1
    else:
        m = 2 * (p + 1)
    factors_dict = factorize(m)
    divisors = generate_divisors(factors_dict)
    for d in divisors:
        if d == 0:
            continue
        f_d, f_d_plus_1 = fib_pair(d, p)
        if f_d % p == 0 and f_d_plus_1 % p == 1:
            return d
    return m  # Fallback, should not reach here

def main():
    import sys
    input = sys.stdin.read().split()
    ptr = 0
    N = int(input[ptr])
    ptr += 1
    factors = []
    for _ in range(N):
        p = int(input[ptr])
        k = int(input[ptr + 1])
        ptr += 2
        factors.append((p, k))
    
    current_lcm = 1
    for (p, k) in factors:
        if p == 2:
            pi_p = 3
        elif p == 5:
            pi_p = 20
        else:
            pi_p = compute_pisano_period(p)
        pi_pk = pi_p * (p ** (k - 1))
        g = math.gcd(current_lcm, pi_pk)
        current_lcm = (current_lcm // g) * pi_pk
        current_lcm %= MOD  # To handle large numbers, mod here is safe since LCM properties
    print(current_lcm % MOD)

if __name__ == "__main__":
    main()
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