結果

問題 No.1516 simple 門松列 problem Re:MASTER
コンテスト
ユーザー convexineq
提出日時 2021-05-21 23:11:01
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 109 ms / 6,000 ms
コード長 3,449 bytes
コンパイル時間 444 ms
コンパイル使用メモリ 82,380 KB
実行使用メモリ 77,508 KB
最終ジャッジ日時 2024-10-10 10:04:30
合計ジャッジ時間 3,164 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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 * 19
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 # 

def mul_poly(f,g):
    df,dg = len(f),len(g)
    res = [0]*(df+dg-1)
    for i in range(df):
        for j in range(dg):
            res[i+j] += f[i]*g[j]
            res[i+j] %= MOD
    return res

def sub_poly(f,g):
    df,dg = len(f),len(g)
    m = max(df,dg)
    res = f[:]+[0]*max(0,m-df)
    for i in range(m):
        if i < dg:
            res[i] -= g[i]
            if res[i] < 0: res[i] += MOD
    return res

def divmod_poly(f,g):
    assert g != [] and g != [0]
    ginv = pow(g[-1],MOD-2,MOD)
    df,dg = len(f),len(g)
    if df < dg: return [0], f[:]

    gg = [gi*ginv%MOD for gi in g]
    r = f[:]
    for i in range(df-dg,-1,-1):
        for j in range(dg-2,-1,-1):
            r[j+i] -= r[dg+i-1]*gg[j]
            r[j+i] %= MOD
    q = r[dg-1:]
    r = r[:dg-1]
    for i in range(df-dg+1): # g を monic にする
        q[i] = q[i]*ginv%MOD
    while r and r[-1]==0: r.pop()
    if not r: r = [0]

    return q,r

"""
数列 f の最小の線形漸化式を発見する
f = r/q (mod X^N) の形で返す
"""
def find_linear_reccurence(f):
    r0 = [0]*len(f)+[1]
    r1 = f[:]
    while r1 and r1[-1]==0: r1.pop()
    if not r1: return [0],[0] # all 0 なら 0 が解
    q0,q1 = [0],[1]
    r_ans, q_ans, size_ans = r1, q1, len(r1)+1

    while r1 != [0]:
        q,r = divmod_poly(r0,r1)
        r0,r1 = r1,r
        q0,q1 = q1,sub_poly(q0,mul_poly(q,q1))
        size = max(len(q1),len(r1)+1)
        if q1[0] and size < size_ans:
            r_ans, q_ans, size_ans = r1, q1, size

    if len(q_ans) <= len(r_ans):
        q_ans += [0]*(len(r_ans)-len(q_ans)+1)
    
    qinv = pow(q_ans[0],MOD-2,MOD)
    for i in range(len(q_ans)): q_ans[i] = q_ans[i]*qinv%MOD
    for i in range(len(r_ans)): r_ans[i] = r_ans[i]*qinv%MOD
    return r_ans, q_ans

def polymul(f,g):
    lf = len(f)
    lg = len(g)
    res = [0]*(lf+lg-1)
    for i in range(lf):
        for j in range(lg):
            res[i+j] += f[i]*g[j]
            res[i+j] %= MOD
    return res

def fps_nth_term(f,g,N):
    assert g[0] != 0
    while N:
        h = g[:]
        for i in range(1,len(g),2):
            h[i] = -h[i]
        f = polymul(f,h)[N%2:N+1:2]
        g = polymul(g,h)[:N+1:2]
        N //= 2
    return f[0]*pow(g[0],MOD-2,MOD)%MOD



n,k = map(int,input().split())
MOD = 998244353

dpn = [[0]*k for _ in range(k)]
dps = [[0]*k for _ in range(k)]
for i in range(k):
    for j in range(i+1,k):
            dpn[i][j] = dpn[j][i] = 1
            dps[i][j] = dps[j][i] = i+j

M = 100
vn = [0]*M
vs = [0]*M
for I in range(M):
    ndpn = [[0]*k for _ in range(k)]
    ndps = [[0]*k for _ in range(k)]
    for i in range(k):
        for j in range(k):
            if i<j:
                for p in range(j):
                    if i==p: continue
                    ndpn[j][p] += dpn[i][j]
                    ndpn[j][p] %= MOD
                    ndps[j][p] += dps[i][j] + dpn[i][j]*p
                    ndps[j][p] %= MOD
            elif i>j:
                for p in range(j+1,k):
                    if i==p: continue
                    ndpn[j][p] += dpn[i][j]
                    ndpn[j][p] %= MOD
                    ndps[j][p] += dps[i][j] + dpn[i][j]*p
                    ndps[j][p] %= MOD

    dpn = ndpn
    dps = ndps
    vn[I] = sum(sum(i) for i in dpn)%MOD
    vs[I] = sum(sum(i) for i in dps)%MOD


r,q = find_linear_reccurence(vn)
print(fps_nth_term(r,q,n-3), end=" ")

r,q = find_linear_reccurence(vs)
print(fps_nth_term(r,q,n-3))
0