結果

問題 No.1346 Rectangle
ユーザー lam6er
提出日時 2025-04-15 21:06:33
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,903 bytes
コンパイル時間 237 ms
コンパイル使用メモリ 82,032 KB
実行使用メモリ 76,840 KB
最終ジャッジ日時 2025-04-15 21:12:41
合計ジャッジ時間 1,955 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
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 AC * 1
other AC * 1 WA * 16
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

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, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        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)
    return sorted(factors)

def is_prime_power(n):
    if n == 1:
        return False
    factors = factor(n)
    return len(set(factors)) == 1

def integer_kth_root(n, k):
    if n <= 0:
        return 0
    if k == 1:
        return n
    low = 1
    high = n
    while low <= high:
        mid = (low + high) // 2
        try:
            powered = pow(mid, k)
        except OverflowError:
            powered = float('inf')
        if powered == n:
            return mid
        elif powered < n:
            low = mid + 1
        else:
            high = mid - 1
    return high

def find_largest_prime_less_than(n):
    if n <= 2:
        return 0
    candidate = n - 1
    while candidate >= 2:
        if is_prime(candidate):
            return candidate
        candidate -= 1
    return 0

def main():
    N = int(input().strip())
    if N == 2:
        print("INF")
        return
    m = N - 1
    if is_prime_power(m):
        print(m)
        return
    max_pp = 0
    max_k = int(math.log2(m)) + 1 if m > 1 else 1
    for k in range(max_k, 1, -1):
        p = integer_kth_root(m, k)
        if p < 2:
            continue
        while True:
            powered = pow(p, k)
            if powered < m:
                break
            p -= 1
            if p < 2:
                break
        if p < 2:
            continue
        if is_prime(p):
            candidate = pow(p, k)
            if candidate > max_pp:
                max_pp = candidate
    largest_prime = find_largest_prime_less_than(m)
    if largest_prime > max_pp:
        max_pp = largest_prime
    print(max_pp)

if __name__ == "__main__":
    main()
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