結果

問題 No.2578 Jewelry Store
ユーザー suisen
提出日時 2023-01-11 19:54:42
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,435 bytes
コンパイル時間 504 ms
コンパイル使用メモリ 82,044 KB
実行使用メモリ 136,832 KB
最終ジャッジ日時 2024-09-27 00:33:24
合計ジャッジ時間 17,992 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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ファイルパターン 結果
other AC * 34 WA * 20
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ソースコード

diff # 

import sys
from typing import List

from math import gcd

input = sys.stdin.readline

# https://qiita.com/Kiri8128/items/eca965fe86ea5f4cbb98
class PrimeFactorize:
    @staticmethod
    def __isPrimeMR(n: int):
        d = n - 1
        d = d // (d & -d)
        L = [2]
        for a in L:
            t = d
            y = pow(a, t, n)
            if y == 1: continue
            while y != n - 1:
                y = (y * y) % n
                if y == 1 or t == n - 1: return 0
                t <<= 1
        return 1

    @staticmethod
    def __findFactorRho(n: int):
        m = 1 << n.bit_length() // 8
        for c in range(1, 99):
            f = lambda x: (x * x + c) % n
            y, r, q, g = 2, 1, 1, 1
            x = ys = y
            while g == 1:
                x = y
                for i in range(r):
                    y = f(y)
                k = 0
                while k < r and g == 1:
                    ys = y
                    for i in range(min(m, r - k)):
                        y = f(y)
                        q = q * abs(x - y) % n
                    g = gcd(q, n)
                    k += m
                r <<= 1
            if g == n:
                g = 1
                while g == 1:
                    ys = f(ys)
                    g = gcd(abs(x - ys), n)
            if g < n:
                if PrimeFactorize.__isPrimeMR(g): return g
                elif PrimeFactorize.__isPrimeMR(n // g): return n // g
                return PrimeFactorize.__findFactorRho(g)

    @staticmethod
    def primeFactor(n):
        i = 2
        ret = {}
        rhoFlg = 0
        while i*i <= n:
            k = 0
            while n % i == 0:
                n //= i
                k += 1
            if k: ret[i] = k
            i += 1 + i % 2
            if i == 101 and n >= 2 ** 20:
                while n > 1:
                    if PrimeFactorize.__isPrimeMR(n):
                        ret[n], n = 1, 1
                    else:
                        rhoFlg = 1
                        j = PrimeFactorize.__findFactorRho(n)
                        k = 0
                        while n % j == 0:
                            n //= j
                            k += 1
                        ret[j] = k

        if n > 1: ret[n] = 1
        if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
        return ret

P = 998244353

t, m = map(int, input().split())
pf = PrimeFactorize.primeFactor(m).keys()
k = len(pf)

m_div_p = [m // p for p in pf]

parity = [bin(s).count('1') & 1 for s in range(1 << k)]

def subset_zeta_product(f: List[int]):
    block = 1
    while block < 1 << k:
        offset = 0
        while offset < 1 << k:
            for i in range(offset, offset + block):
                f[i + block] = f[i + block] * f[i] % P
            offset += 2 * block
        block <<= 1

def solve():
    _, x0, c, d = map(int, input().split())

    prod = [1] * (1 << k)

    wi = x0
    for ai in map(int, input().split()):
        if m % ai == 0:
            t = 0
            for j, mp in enumerate(m_div_p):
                t |= (mp % ai != 0) << j
            prod[t] = prod[t] * (1 + wi) % P
        wi = (c * wi + d) % P
    
    subset_zeta_product(prod)

    ans = 0
    for s in range(1 << k):
        if parity[s]:
            ans -= prod[s]
        else:
            ans += prod[s]
    if m == 1:
        ans -= 1

    print(ans % P)

for _ in range(t):
    solve()
0