結果

問題 No.1387 Mitarushi's Remodeling
ユーザー Mitarushi
提出日時 2021-01-19 22:53:15
言語 PyPy3
(7.3.15)
結果
RE  
(最新)
AC  
(最初)
実行時間 -
コード長 1,836 bytes
コンパイル時間 188 ms
コンパイル使用メモリ 82,432 KB
実行使用メモリ 235,356 KB
最終ジャッジ日時 2024-06-29 22:59:47
合計ジャッジ時間 55,065 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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 * 3
other AC * 11 RE * 56
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

mod = 998244353
max_kai = 4*10**5
kai = [1]
for i in range(1, max_kai):
    kai.append(kai[-1] * i % mod)
kkai = [0] * max_kai
kkai[-1] = pow(kai[-1], mod - 2, mod)
for i in reversed(range(1, max_kai)):
    kkai[i - 1] = kkai[i] * i % mod
kkai_sum = [0]*(max_kai+1)
for i in range(max_kai):
    kkai_sum[i+1] = (kkai_sum[i] + kkai[i]) % mod
def cutsum(a, b):
    return kkai_sum[a] - kkai_sum[max(0, b)]
def fft_inplace(a, w):
    n = len(a)
    m = n
    t = 1
    while m >= 2:
        mh = m >> 1
        for i in range(0, n, m):
            for s in range(mh):
                j, k = i+s, i+mh+s
                a[j], a[k] = (a[j]+a[k]) % mod, (a[j]-a[k])*w[s*t] % mod
        m = mh
        t *= 2
def ifft_inplace(a, w):
    n = len(a)
    m = 2
    t = -(n >> 1)
    while m <= n:
        mh = m >> 1
        for i in range(0, n, m):
            for s in range(mh):
                j, k = i+s, i+mh+s
                a[k] *= w[s*t]
                a[j], a[k] = (a[j]+a[k]) % mod, (a[j]-a[k]) % mod
        m <<= 1
        t //= 2
    n_inv = pow(n, mod-2, mod)
    for i in range(n):
        a[i] = a[i] * n_inv % mod
def convolution(a, b):
    n2 = max(len(a) + len(b), 1)
    n = 1 << (n2-1).bit_length()
    a = a + [0] * (n-len(a))
    b = b + [0] * (n-len(b))
    w_root = pow(3, (mod-1)//n, mod)
    w = [1] * n
    for i in range(1, n):
        w[i] = w[i-1] * w_root % mod
    fft_inplace(a, w)
    fft_inplace(b, w)
    c = [i*j % mod for i, j in zip(a, b)]
    ifft_inplace(c, w)
    return c[:n2]
n, k = map(int, input().split())
m = [int(i)+1 for i in input().split()]
b = convolution(m, m[::-1])
ans = 0
for i in range(n):
    ans += b[i] * (kai[n-1] * cutsum(n-1, n-k-1) - kai[i]*cutsum(i, i-k))
    ans %= mod
for i in m:
    ans *= i-1
    ans %= mod
ans *= pow(4, mod-2, mod)
ans %= mod
print(ans)
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