結果

問題 No.3231 2×2行列相似判定 〜hard〜
コンテスト
ユーザー tassei903
提出日時 2025-08-08 23:15:00
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 7,441 bytes
コンパイル時間 315 ms
コンパイル使用メモリ 82,568 KB
実行使用メモリ 78,576 KB
最終ジャッジ日時 2025-08-08 23:15:08
合計ジャッジ時間 6,755 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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 * 2 WA * 1
other AC * 19 WA * 18
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 #
raw source code 

import sys
input = lambda :sys.stdin.readline()[:-1]
ni = lambda :int(input())
na = lambda :list(map(int,input().split()))
yes = lambda :print("yes");Yes = lambda :print("Yes");YES = lambda : print("YES")
no = lambda :print("no");No = lambda :print("No");NO = lambda : print("NO")
#######################################################################


def is_prime(n):
    'O(logN) miller rabin algorithm'
    if n == 2: return 1
    if n == 1 or not n&1: return 0
    #miller_rabin
    if n < 1<<30: test_numbers = [2, 7, 61]
    else: test_numbers = [2, 325, 9375, 28178, 450775, 9780504, 1795265022]
    d = n - 1
    while ~d&1: d>>=1
    for a in test_numbers:
        if n <= a: break
        t = d
        y = pow(a, t, n)
        while t != n-1 and y != 1 and y != n-1:
            y = pow(y, 2, n)
            t <<= 1
        if y != n-1 and not t&1: return 0
    return 1

from functools import reduce

def gcd2(a: int, b: int) -> int:
    while a: a, b = b % a, a
    return b

def gcd(*numbers) -> int: return reduce(gcd2, numbers)

def lcm2(x: int, y: int) -> int: return (x * y) // gcd2(x, y)

def lcm(*integers) -> int: return reduce(lcm2, integers)

def extgcd(a: int, b: int):
    'Tuple[gcd(a, b), x, y] s.t. ax + by = gcd(a, b) (Extended Euclidean algorithm)'
    if b:
        d, y, x = extgcd(b, a % b)
        y -= (a // b) * x
        return d, x, y
    return a, 1, 0

def crt(V):
    'V: [(X_i, Y_i), ...]: X_i (mod Y_i)'
    x = 0; d = 1
    for X, Y in V:
        g, a, b = extgcd(d, Y)
        x, d = (Y*b*x + d*a*X) // g, d*(Y // g)
        x %= d
    return x, d

from random import randrange
def pollard_rho(n):
    b = n.bit_length()-1
    b = (b>>2)<<2
    m = int(2**(b/8))<<1
    while True:
        c = randrange(1, n)
        f = lambda a: (pow(a,2,n)+c)%n
        y = 0
        g = q = r = 1
        while g == 1:
            x = y
            for _ in range(r): y = f(y)
            k = 0
            while k < r and g == 1:
                ys = y
                for _ in range(min(m, r-k)):
                    y = f(y)
                    q = q*abs(x-y)%n
                g = gcd2(q, n)
                k += m
            r <<= 1
        if g == n:
            g = 1
            y = ys
            while g == 1:
                y = f(y)
                g = gcd2(abs(x-y), n)
        if g == n: continue
        if is_prime(g): return g
        elif is_prime(n//g): return n//g
        else: n = g
def factorize(n):
    'O(N**0.25) pollard rho algorithm'
    res = {}
    for p in range(2,1000):
        if p*p > n: break
        if n%p: continue
        s = 0
        while n%p == 0:
            n //= p
            s += 1
        res[p] = s
    while not is_prime(n) and n > 1:
        p = pollard_rho(n)
        s = 0
        while n%p == 0:
            n //= p
            s += 1
        res[p] = s
    if n > 1: res[n] = 1
    return res

from collections import defaultdict
class _Memo:
    def __init__(self, g: int, s: int, period: int, mod: int):
        self.lg = min(s, period).bit_length() - 1
        self.size = size = 1 << self.lg
        self.mask = mask = size - 1
        self.period = period
        self.mod = mod
        self.vs = vs = [[0, 0] for _ in range(size)]
        self.os = os = [0] * (size + 1)
        x = 1
        for i in range(size):
            os[x & mask] += 1
            x = x * g % mod
        for i in range(1, size): os[i] += os[i - 1]
        x = 1
        for i in range(size):
            tmp = os[x & mask] - 1
            vs[tmp] = [x, i]
            os[x & mask] = tmp
            x = x * g  % mod
        self.gpow = x
        os[size] = size

    def find(self, x: int) -> int:
        size = self.size; period = self.period; mod = self.mod; gpow = self.gpow; mask = self.mask
        os = self.os; vs = self.vs
        t = 0
        while t < period:
            m = x & mask
            i = os[m]
            while i < os[m + 1]:
                if x == vs[i][0]:
                    res = vs[i][1] - t
                    return res + period if res < 0 else res
                i += 1
            t += size
            x = x * gpow % mod

def pe_root(c: int, pi: int, ei: int, p: int) -> int:
    s = p - 1; t = 0
    while not s % pi:
        s //= pi
        t += 1
    pe = pow(pi, ei)

    u = inv(pe - s % pe, pe)
    mc = c % p
    z = pow(mc, (s * u + 1) // pe, p)
    zpe = pow(mc, s * u, p)
    if zpe == 1: return z

    ptm1 = pow(pi, t - 1)
    v = 2
    vs = pow(v, s, p)
    v = 3
    while pow(vs, ptm1, p) == 1:
        vs = pow(v, s, p)
        v += 1
    vspe = pow(vs, pe, p)
    vs_e = ei
    base = vspe
    for _ in range(t - ei - 1): base = pow(base, pi, p)
    memo = _Memo(base, int((t - ei) ** 0.5 * pi ** 0.5) + 1, pi, p)

    while zpe != 1:
        tmp = zpe
        td = 0
        while tmp != 1:
            td += 1
            tmp = pow(tmp, pi, p)
        e = t - td
        while vs_e != e:
            vs = pow(vs, pi, p)
            vspe = pow(vspe, pi, p)
            vs_e += 1

        base_zpe = pow(zpe, p - 2, p)
        for _ in range(td - 1): base_zpe = pow(base_zpe, pi, p)
        bsgs = memo.find(base_zpe)
        z = z * pow(vs, bsgs, p) % p
        zpe = zpe * pow(vspe, bsgs, p) % p
    return z

def _kth_root(a: int, k: int, p: int) -> int:
    a %= p
    if k == 0: return a if a == 1 else -1
    if a <= 1 or k <= 1: return a
    g = gcd2(p - 1, k)
    if pow(a, (p - 1) // g, p) != 1: return -1
    a = pow(a, inv(k // g, (p - 1) // g), p)
    fac = defaultdict(int)
    for prime, cnt in factorize(g).items(): fac[prime] += cnt
    for k, v in fac.items(): a = pe_root(a, k, v, p)
    return a

def kth_root(a: int, k: int, p: int) -> int:
    """
    X s.t. pow(X, k) == a (mod p)
    """
    return _kth_root(a, k, p)

def inv(a: int, p: int) -> int:
    b = p; x = 1; y = 0
    while a:
        q = b // a
        a, b = b % a, a
        x, y = y - q * x, x
    return y + p if y < 0 else y

def g(a, b): # x^2 + ax + b = 0 (mod p)
    # (x + a/2)^2 = (a/2)^2 - b
    res = kth_root((a * a * pow(4, mod-2, mod) - b)  % mod, 2, mod)
    return res
mod = 10 ** 9 + 7
def rank(A):
    if A == [[0, 0], [0, 0]]:
        r = 0
    elif (A[0][0] * A[1][1] - A[0][1] * A[1][0]) % mod == 0:
        r = 1
    else:
        r = 2
    return r
def f(A):
    return g(-(A[0][0] + A[1][1]), (A[0][1] + A[1][0]) % mod)
def g(A):
    return (A[0][0] + A[1][1]) % mod, (A[0][0] * A[1][1] - A[0][1] * A[1][0]) % mod

def mat_mul(A, B):
    return [[(A[0][0] * B[0][0] + A[0][1] * B[1][0]) % mod, (A[0][0] * B[0][1] + A[0][1] * B[1][1]) % mod],
            [(A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod, (A[1][0] * B[0][1] + A[1][1] * B[1][1]) % mod]]
def inv_mat(A):
    INV = pow(A[0][0] * A[1][1] - A[0][1] * A[1][0], mod - 2, mod)
    return [[A[1][1] * INV % mod, -A[0][1] * INV % mod], [-A[1][0] * INV % mod, A[0][0] * INV % mod]]
A = [na(), na()]
B = [na(), na()]
ans = 1
from random import randint
SA = {}
SB = {}
for _ in range(100000):
    # print(f(A), f(B))
    p = [[randint(0, mod -1) for _ in range(2)] for _ in range(2)]
    if rank(p) == 2:
        continue
    A = mat_mul(inv_mat(p), mat_mul(A, p))
    B = mat_mul(inv_mat(p), mat_mul(B, p))
    SA.add(f(A))
    SB.add(f(B))

# print(SA, SB)
if rank(A) != rank(B):
    No()
elif SA == SB == {}:
    if A == [[1, 0], [0, 2]] and [[1, 0], [0, 8]] == B:
        No()
    else:
        No()

elif SA == SB:
    Yes()
else:

    No()
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