結果

問題 No.1753 Don't cheat.
ユーザー zkouzkou
提出日時 2021-06-19 17:04:36
言語 Python3
(3.12.2 + numpy 1.26.4 + scipy 1.12.0)
結果
TLE  
実行時間 -
コード長 2,812 bytes
コンパイル時間 240 ms
コンパイル使用メモリ 13,184 KB
実行使用メモリ 40,136 KB
最終ジャッジ日時 2024-12-31 18:23:38
合計ジャッジ時間 101,184 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 32 ms
17,696 KB
testcase_01 AC 33 ms
35,712 KB
testcase_02 AC 30 ms
17,700 KB
testcase_03 AC 30 ms
35,968 KB
testcase_04 AC 29 ms
17,700 KB
testcase_05 AC 33 ms
36,096 KB
testcase_06 AC 31 ms
18,084 KB
testcase_07 TLE -
testcase_08 TLE -
testcase_09 TLE -
testcase_10 TLE -
testcase_11 TLE -
testcase_12 TLE -
testcase_13 TLE -
testcase_14 TLE -
testcase_15 TLE -
testcase_16 TLE -
testcase_17 TLE -
testcase_18 TLE -
testcase_19 TLE -
testcase_20 TLE -
testcase_21 TLE -
testcase_22 TLE -
testcase_23 TLE -
testcase_24 TLE -
testcase_25 TLE -
testcase_26 TLE -
testcase_27 TLE -
testcase_28 TLE -
testcase_29 TLE -
testcase_30 TLE -
testcase_31 TLE -
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ソースコード

diff # 

MOD = 998244353
half = pow(2, MOD - 2, MOD)


def int2frac(n, m=None, N = 10000, D = 10000):
    """
    Return (r, s) s.t. r = s * n (mod m) if such a pair exists.
    Otherwise, return (0, 0).

    Parameters
    ----------
    n: an integer that will be represented as a fraction.   
    m: A modulus used to represent n.     
    N: An upperbound of r, which is the numerator of n. 
    D: An upperbound of s, which is the denominator of n.    
    """
    def gcd(a, b):
        while b:
            a, b = b, a % b
        return a
    if m is None:
        m = MOD
    v = (m, 0)
    w = (n, 1)
    while w[0] > N:
        q = v[0] // w[0]
        v, w = w, (v[0] - q * w[0], v[1] - q * w[1])
    if w[1] < 0:
        w = (-w[0], -w[1])
    if w[1] <= D and gcd(w[0], w[1]) == 1:
        return w
    else:
        return (0, 0)


def fwht(a) -> None:
    """
    In-place Fast Walsh–Hadamard Transform of array a.
    Reference: https://en.wikipedia.org/wiki/Fast_Walsh%E2%80%93Hadamard_transform
    """
    h = 1
    while h < len(a):
        for i in range(0, len(a), h * 2):
            for j in range(i, i + h):
                x = a[j]
                y = a[j + h]
                a[j] = (x + y) % MOD
                a[j + h] = (x - y) % MOD
        h *= 2

def ifwht(a) -> None:
    """
    In-place Inverse Fast Walsh–Hadamard Transform of array a.
    Reference: https://en.wikipedia.org/wiki/Fast_Walsh%E2%80%93Hadamard_transform
    """
    h = 1
    while h < len(a):
        for i in range(0, len(a), h * 2):
            for j in range(i, i + h):
                x = a[j]
                y = a[j + h]
                a[j] = (x + y) * half % MOD
                a[j + h] = (x - y) * half % MOD
        h *= 2

N = int(input())
As = list(map(int, input().split()))

assert 1 <= N <= 2 * 10 ** 3
assert all(0 <= A <= 10 ** 5 for A in As)
assert N + 1 == len(As)
assert As[0]

sumAs = sum(As)
invsumAs = pow(sum(As), MOD - 2, MOD)
for i in range(N + 1):
    As[i] *= invsumAs
    As[i] %= MOD

# print([int2frac(A) for A in As])

z = 1 << N.bit_length()

A0 = [0] * z
A0[0] = As[0]
fwht(A0)
As_fwht = As + [0] * (z - N - 1)
fwht(As_fwht)
assert all((As[0] + 1 - As_fwht[k]) % MOD != 0 for k in range(z))
q = [pow(As[0] + 1 - As_fwht[k], MOD - 2, MOD) * A0[k] % MOD for k in range(z)]
ifwht(q)

ps = []
for x in range(1, N + 1):
    A0 = [0] * z
    A0[0] = As[x]
    fwht(A0)
    Ax = As + [0] * (z - N - 1)
    Ax[0] = 0
    Ax[x] = 0
    fwht(Ax)
    assert all((1 - Ax[k]) % MOD != 0 for k in range(z))
    p = [pow(1 - Ax[k], MOD - 2, MOD) * A0[k] % MOD for k in range(z)]
    ifwht(p)
    ps.append(p)
    # print(x, [int2frac(e) for e in p])

answer = q[0]
for p in ps:
    for xor in range(z):
        answer += p[xor] * q[xor] % MOD
        answer %= MOD

print((1 - answer) % MOD)
0