結果

問題 No.3589 Make Ends Meet (Hard)
コンテスト
ユーザー marc2825
提出日時 2026-05-29 17:03:07
言語 PyPy3
(7.3.17)
コンパイル:
pypy3 -mpy_compile _filename_
実行:
pypy3 _filename_
結果
AC  
実行時間 801 ms / 2,000 ms
コード長 3,231 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 227 ms
コンパイル使用メモリ 95,852 KB
実行使用メモリ 272,264 KB
最終ジャッジ日時 2026-07-10 20:55:38
合計ジャッジ時間 9,606 ms
ジャッジサーバーID
（参考情報） 		judge1_0 / judge3_0
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 * 47
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 

MOD = 998244353

def modpow(a, e):
    r = 1
    while e > 0:
        if e & 1:
            r = r * a % MOD
        a = a * a % MOD
        e >>= 1
    return r

def C(n, r):
    if r < 0 or r > n:
        return 0
    return fact[n] * ifact[r] % MOD * ifact[n - r] % MOD

def mul_qs_minus_1_full(p, s, E):
    res = [0] * (E + 1)

    # p(q) * q^s
    if s <= E:
        for i in range(E + 1 - s):
            res[i + s] = p[i]

    # - p(q)
    for i in range(E + 1):
        res[i] -= p[i]
        if res[i] < 0:
            res[i] += MOD

    return res

def add_shift_scaled_full(dst, src, shift, scale, E):
    if scale == 0 or shift > E:
        return

    for i in range(E + 1 - shift):
        if src[i] == 0:
            continue
        dst[i + shift] = (dst[i + shift] + src[i] * scale) % MOD


N, M, K = map(int, input().split())

E = N * (N - 1) // 2
R = E - M
S = N - 2

fact = [1] * (E + 1)
ifact = [1] * (E + 1)

for i in range(1, E + 1):
    fact[i] = fact[i - 1] * i % MOD

ifact[E] = modpow(fact[E], MOD - 2)

for i in range(E, 0, -1):
    ifact[i - 1] = ifact[i] * i % MOD

# dp[r][s] is a polynomial in q of fixed length E+1.
# r: number of ordinary vertices not reached yet
# s: size of current BFS frontier
dp = [[[0] * (E + 1) for _ in range(S + 2)] for _ in range(S + 1)]
active = [[False] * (S + 2) for _ in range(S + 1)]

dp[S][1][0] = 1
active[S][1] = True

# Build layers 1,2,...,K-1 without reaching vertex N.
for step in range(K - 1):
    ndp = [[[0] * (E + 1) for _ in range(S + 2)] for _ in range(S + 1)]
    nactive = [[False] * (S + 2) for _ in range(S + 1)]

    for r in range(S + 1):
        for s in range(S + 2):
            if not active[r][s]:
                continue

            cur = dp[r][s][:]  # cur * (q^s - 1)^t

            for t in range(r + 1):
                shift = t * (t - 1) // 2
                ways = C(r, t)

                add_shift_scaled_full(ndp[r - t][t], cur, shift, ways, E)
                nactive[r - t][t] = True

                if t != r:
                    cur = mul_qs_minus_1_full(cur, s, E)

    dp = ndp
    active = nactive

# H(q): generating polynomial for dist(1,N)=K in q=1+x.
H = [0] * (E + 1)

# N must have at least one edge to the current frontier: (q^s - 1).
# All edges among the remaining r ordinary vertices, between them and N,
# and between them and the current frontier are free.
for r in range(S + 1):
    for s in range(S + 2):
        if not active[r][s]:
            continue

        p = dp[r][s]
        base = s * r + r + r * (r - 1) // 2

        # p(q) * (q^s - 1) * q^base
        for i in range(E + 1):
            val = p[i]
            if val == 0:
                continue

            # + p(q) * q^{base+s}
            if i + base + s <= E:
                H[i + base + s] += val
                if H[i + base + s] >= MOD:
                    H[i + base + s] -= MOD

            # - p(q) * q^base
            if i + base <= E:
                H[i + base] -= val
                if H[i + base] < 0:
                    H[i + base] += MOD

# We need [x^R] H(1+x).
# If H(q)=sum_c h_c q^c, then
# [x^R] H(1+x) = sum_c h_c * C(c,R).
ans = 0

for c in range(R, E + 1):
    ans += H[c] * C(c, R)
    ans %= MOD

print(ans)
0