結果

問題 No.847 Divisors of Power
ユーザー lam6er
提出日時 2025-03-20 21:15:37
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 91 ms / 2,000 ms
コード長 1,532 bytes
コンパイル時間 166 ms
コンパイル使用メモリ 82,720 KB
実行使用メモリ 76,200 KB
最終ジャッジ日時 2025-03-20 21:16:23
合計ジャッジ時間 2,555 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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ファイルパターン 結果
sample AC * 4
other AC * 26
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 #
プレゼンテーションモードにする

def factorize(n):
    factors = []
    if n == 1:
        return factors
    i = 2
    while i * i <= n:
        if n % i == 0:
            cnt = 0
            while n % i == 0:
                cnt += 1
                n = n // i
            factors.append((i, cnt))
        i += 1
    if n > 1:
        factors.append((n, 1))
    return factors
def count_divisors(primes, M, index, current_prod):
    if current_prod > M:
        return 0
    if index == len(primes):
        return 1
    total = 0
    p, e_max = primes[index]
    quotient = M // current_prod
    if quotient == 0:
        return 0
    # Compute maximum exponent a such that p^a <= quotient
    max_power = 0
    current_p = 1
    while True:
        if current_p > quotient:
            break
        max_power += 1
        current_p *= p
    max_power -= 1
    max_a = min(max_power, e_max)
    # Now iterate through possible exponents
    current_p = 1
    for a in range(0, max_a + 1):
        new_prod = current_prod * current_p
        if new_prod > M:
            break
        total += count_divisors(primes, M, index + 1, new_prod)
        current_p *= p
    return total
def main():
    import sys
    input_line = sys.stdin.readline()
    N, K, M = map(int, input_line.strip().split())
    if N == 1:
        print(1 if M >= 1 else 0)
        return
    factors = factorize(N)
    primes_K = [(p, e * K) for (p, e) in factors]
    primes_K.sort()
    result = count_divisors(primes_K, M, 0, 1)
    print(result)
if __name__ == "__main__":
    main()
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