結果

問題 No.577 Prime Powerful Numbers
ユーザー gew1fw
提出日時 2025-06-12 16:25:13
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 4,069 bytes
コンパイル時間 433 ms
コンパイル使用メモリ 82,688 KB
実行使用メモリ 101,888 KB
最終ジャッジ日時 2025-06-12 16:25:20
合計ジャッジ時間 6,927 ms
ジャッジサーバーID
（参考情報） 		judge1 / 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 -- * 1
other TLE * 1 -- * 9
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

def sieve(limit):
    sieve = [True] * (limit + 1)
    sieve[0] = sieve[1] = False
    for i in range(2, int(math.sqrt(limit)) + 1):
        if sieve[i]:
            for j in range(i*i, limit+1, i):
                sieve[j] = False
    primes = [i for i, is_prime in enumerate(sieve) if is_prime]
    return primes

# Precompute primes up to 1e6
primes_list = sieve(10**6)

def is_prime(n):
    if n < 2:
        return False
    for p in primes_list:
        if p * p > n:
            break
        if n % p == 0:
            return False
    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 integer_nth_root(n, b):
    if n == 0:
        return 0
    low = 1
    high = n
    while low <= high:
        mid = (low + high) // 2
        powered = 1
        overflow = False
        for _ in range(b):
            powered *= mid
            if powered > n:
                overflow = True
                break
        if overflow:
            high = mid - 1
        elif powered == n:
            return mid
        elif powered < n:
            low = mid + 1
        else:
            high = mid - 1
    powered = 1
    for _ in range(b):
        powered *= high
        if powered > n:
            break
    if powered == n:
        return high
    else:
        return -1

def is_prime_power(n):
    if n < 2:
        return False
    for b in range(1, 61):
        x = integer_nth_root(n, b)
        if x == -1:
            continue
        if x ** b == n and is_prime(x):
            return True
    return False

def main():
    input = sys.stdin.read().split()
    Q = int(input[0])
    for i in range(1, Q+1):
        N = int(input[i])
        found = False

        # Handle a=1 case: p is a prime, rem = N - p must be a prime
        # Alternatively, rem = N - q^b must be a prime
        # So, loop through all possible q^b and check if rem is prime
        for b in range(1, 61):
            for q in primes_list:
                q_power = q ** b
                if q_power > N - 2:
                    break
                rem = N - q_power
                if rem < 2:
                    continue
                if is_prime(rem):
                    print("Yes")
                    found = True
                    break
            if found:
                break
        if found:
            continue

        # Handle a >=2 case: p^a is part, rem must be a prime power
        for a in range(2, 61):
            if a > 60:
                break
            max_p = int((N - 2) ** (1.0 / a))
            for p in primes_list:
                if p > max_p:
                    break
                p_power = p ** a
                rem = N - p_power
                if rem < 2:
                    continue
                if is_prime_power(rem):
                    print("Yes")
                    found = True
                    break
            if found:
                break
        if found:
            continue

        # Check the other way around for a=1 (when p is a prime and rem is q^b)
        # This part is similar to the a=1 case but we have to do it again as the previous loop may not cover all possibilities
        for b in range(1, 61):
            for q in primes_list:
                q_power = q ** b
                if q_power > N - 2:
                    break
                rem = N - q_power
                if rem < 2:
                    continue
                if is_prime_power(rem):
                    print("Yes")
                    found = True
                    break
            if found:
                break
        if found:
            continue

        print("No")

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